MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 3

Find the Lactus Rectum of

$\displaystyle 9y^{2}-4x^{2}=36$

0%
$9$
0%
$6$
0%
$11$
0%
$15$

Explanation

$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$
Here the coefficient of

$\displaystyle y^{2}$ is + ive and that of

$\displaystyle x^{2}$ is -ive and hence it represents a hyperbola whose transerse axis is vertical, i.e.

$\displaystyle a^{2}=4, b^{2}=9.$

$\displaystyle b^{2}= a^{2}\left ( e^{2}-1 \right )$
or

$\displaystyle \frac{9}{4}+1=e^{2}\therefore e= \frac{\sqrt{13}}{2}$
Foci lie on y-axis

$\displaystyle \left ( 0, \pm ae \right )$ i.e

$\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$

$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$

The equation of a diameter of a circle is

$x+y=1$ and the greatest distance of any point of the circle from the diameter is

$\dfrac{1}{\sqrt{2}}$ .Then, a possible equation of the circle can be

0%
$x^{2}+y^{2}-2x+4y=0$
0%
$x^{2}+y^{2}-x-y=0$
0%
$x^{2}+y^{2}+4x-2y=0$
0%
$x^{2}+y^{2}+2x+4y=0$

Explanation

Given diameter of circle is

$x+y=1$
Diameter crosses the x-axis at

$(1,0)$ and y-axis at

$(0,1)$
Distance between these two points i.e.

$d=\sqrt{2}$

$\Rightarrow \displaystyle r=\frac{\sqrt{2}}{2}$
Center of the circle is

$\displaystyle (\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } )$
Equation of circle is

$\displaystyle { (x-\frac { 1 }{ 2 } ) }^{ 2 }+{ (y-\frac { 1 }{ 2 } ) }^{ 2 }=\frac { 1 }{ 2 }$

$x^{2}+y^{2}-x-y=0$

Two vertices of an equilateral triangle are

$(-1, 0)$ and

$(1, 0)$ , and its third vertex lies above the

$x$ -axis. The equation of the circumcircle of the triangle is

0%
$x^{2}+y^{2}=1$
0%
$\sqrt{3}\left ( x^{2}+y^{2} \right )+2y-\sqrt{3}=0$
0%
$\sqrt{3}\left ( x^{2}+y^{2} \right )-2y-\sqrt{3}=0$
0%
none of these

Explanation

Points

$A(-1,0)$ and

$B(1,0)$ lie on

$x$ -axis.

Mid-point of

$AB$ is

$(0,0)$ .

Clearly, point

$C$ lies on

$y$ -axis. Since, it lies above

$y$ -axis.

So, let the coordinates of point

$C$ be

$(0, k)$ .

Coordinates of centroid

$G$ is

$\displaystyle \left (0,\dfrac{k}{3}\right)$ .
We also know that circumcenter, centroid and incenter of the equilateral triangle coincides.

Hence, the coordinates of circumcenter is

$\displaystyle \left (0,\dfrac{k}{3}\right)$ .

Since,

$ABC$ is equilateral triangle,

$AB=AC$

$\Rightarrow (AB)^2=(AC)^{2}$

$\Rightarrow k=\sqrt{3}$

So, the coordinates of circumcenter is

$\displaystyle \left (0,\dfrac{1}{\sqrt{3}}\right)$ .

Radius

$CG=\displaystyle \dfrac{2}{\sqrt{3}}$

Equation of circumcircle is

$(x-0)^{2}+\left (y-\dfrac{1}{\sqrt{3}}\right)^{2}=\displaystyle \dfrac{4}{3}$

$\Rightarrow x^{2}+y^2-\dfrac{2}{\sqrt{3}}y-1=0$

$\Rightarrow \displaystyle \sqrt{3}\left ( x^{2}+y^{2} \right )-2y-\sqrt{3}=0$

The equation

$\displaystyle 3x^{2}-2xy+y^{2}=0$ represents:

0%
a circle
0%
hyperbola
0%
a pair of lines
0%
none of these

Explanation

Given expression,

$\displaystyle 3{ x }^{ 2 }-2xy+{ y }^{ 2 }=0$
As Coefficient of

$\displaystyle xy$ is not zero,It will not be a circle and hyperbola.
Let

$\displaystyle \frac { y }{ x } =m$

We get

$\displaystyle { m }^{ 2 }-2m+3=0$ will not have any real solutions as discriminant is less than zero.

$\displaystyle \therefore$ They will not be pair of lines too.

Find the length of latus rectum of the parabola whose focus is the point

$\displaystyle \left ( 2,3 \right )$ and directrix is the line

$\displaystyle x-4y+3=0.$

0%
$\dfrac{14}{\sqrt { 17 }}$
0%
$\dfrac{17}{\sqrt { 17 }}$
0%
$\dfrac{15}{\sqrt { 17 }}$
0%
$\dfrac{10}{\sqrt { 17 }}$

Explanation

By definition the distance of any point on the parabola from the focus is equal to its distance from the directrix.

$\displaystyle \therefore \sqrt{\left \{ \left ( x-2 \right )^{2}+\left ( x-3 \right )^{2} \right \}}=\frac{x-4y+3}{\left ( 1+16 \right )^{1/2}},$

$\displaystyle 17\left ( x^{2}+y^{2}-4x-6y+13 \right )=x^{2}+16y^{2}+9 -8xy-24y+6x$
or

$\displaystyle 16x^{2}+y^{2}+8xy-74x-78y+212=0$
We know L.R.

$\displaystyle =4a=2\left ( 2a \right )$ where

$2a$ is the distance between the focus and directrix

$\displaystyle =2\left|\frac{2.1-4.3+3}{\left ( 1+16 \right )^{1/2}}\right|=\frac{14}{\sqrt{17}}.$

Ans: A

The length of the latus rectum of the parabola

$x=ay^2+by+c$ is

0%
$\displaystyle \frac{a}{4}$
0%
$\displaystyle \frac{a}{3}$
0%
$\displaystyle \frac{1}{a}$
0%
$\displaystyle \frac{1}{4a}$

Explanation

Given,

$x=ay^2+by+c$

$ay^2+by=x-c$

$a\left ( y+\dfrac{b}{a} \right )^2=x+\dfrac{ab^2}{4}-c$

$\left ( y+\dfrac{b}{a} \right )^2=\dfrac{1}{a}\left ( x+\dfrac{ab^2}{4}-c \right )$

Length of latus rectum

$=\dfrac{1}{a}$

On the line joining the points

$A (0,4)$ and

$B (3, 0)$ , a square

$ABCD$ is constructed on the side of the line away from the origin. Equation of the circle having centre at

$C$ and touching the axis of

$x$ is

0%
$x^{2} + y^{2} - 14x -6y + 49 = 0$
0%
$x^{2} + y^{2}- 14x -6y + 9 = 0$
0%
$x^{2} + y^{2}-6x -14y + 49 = 0$
0%
$x^{2} + y^{2}-6x -14y + 9 = 0$

Explanation

Let

$\angle ABO = \theta$ , then

$\angle CBL= 90^{\circ} -\theta, CL$ being perpendicular to x-axis

$(Fig)$ .

The coordinates of

$C$ are

$(OL, LC)$

$OL = OB + BL= 3 + 5 \sin \theta$

$= 3 + 5 \times \left (\dfrac 45\right) = 7$

$CL = 5 \cos \theta = 5 \times \left (\dfrac 35\right) =3$
So, the coordinate of

$C$ are

$(7, 3)$ and the equation of the circle having

$C$ as centre and touching

$x$ -axis is

$(x -7)^{2} + (y-3)^{2} = (CL)^{2} = 9$

$\Rightarrow x^{2} + y^{2} -14x -6y + 49 = 0$

If the lines

$2x+3y+1= 0$ and

$3x-y-4= 0$ lie along the diameter of a circle of circumference

$10\pi$ , then equation of circle be

0%
$x^{2}+y^{2}+2\left ( x+y \right )-23= 0$
0%
$x^{2}+y^{2}-2\left ( x+y \right )-23= 0$
0%
$x^{2}+y^{2}-2x+2y-23= 0$
0%
$x^{2}+y^{2}+2x-2y-23= 0$

Explanation

Given lines

$2x+3y+1= 0$ and

$3x-y-4= 0$
Point of intersection of these lines is the center of circle.
So, center is at (1,-1).

Also, given circumference

$=10\pi$

$\Rightarrow 2\pi r=10\pi$

$\Rightarrow r=5$

Hence the equation of circle is

$(x-1)^2 +(y+1)^2=25$

$x^2+y^2-2x+2y-23=0$

The equation of circle with origin as center and passing through the vertices of an equilateral triangle whose median is of length

$3a$ is

0%
${ x }^{ 2 }+{ y }^{ 2 }=9{ a }^{ 2 }$
0%
${ x }^{ 2 }+{ y }^{ 2 }=16{ a }^{ 2 }$
0%
${ x }^{ 2 }+{ y }^{ 2 }=4{ a }^{ 2 }$
0%
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$

Explanation

The centroid of an equilateral triangle is the center of its circumcenter and the radius of the circle is the distance of any vertex from the centroid i.e. radius of the circle

$=$ distance of centroid from any vertex

$\displaystyle =\frac { 2 }{ 3 } ($ Median

$)\displaystyle =\frac { 2 }{ 3 } (3a)=2a$

Hence, equation of circle whose center is

$(0,0)$ and radius

$2a$ is

${ \left( x-0 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left( 2a \right) }^{ 2 }\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=4{ a }^{ 2 }$

The lines

$2x -3y=5$ and

$3x -4y =7$ are the diameters of a circle of area

$154$ square units. An equation of this circle is

$(\pi = 22/7)$

0%
$x^{2} + y^{2} + 2x -2y = 62$
0%
$x^{2} + y^{2} + 2x -2y = 47$
0%
$x^{2}+y^{2}-2x + 2y = 47$
0%
$x^{2} + y^{2} -2x + 2y = 62$

Explanation

The centre of the circle is the point of intersection of the given diameters

$2x- 3y = 5$ and

$3x- 4y= 7,$ which is

$(1,-1)$
and if

$r$ is the radius of the circle,then

$\displaystyle \pi r^{2} = 154 \Rightarrow r^{2} = 154 \times \frac{7}{22} \Rightarrow r = 7.$
Hence equation of the required

circle is

$(x-1)^{2}+(y+ 1)^{2}=7^{2} \Rightarrow x^{2}+y^{2}- 2x+2y=47$

The equation of circle with origin as a centre and passing through equilateral triangle whose median is of length

$3a$ is

0%
$x^{2} + y^{2} = 9a^{2}$
0%
$x^{2} + y^{2} = 16a^{2}$
0%
$x^{2} + y^{2} = 4a^{2}$
0%
$x^{2} + y^{2} = a^{2}$

Explanation

Given, median of the equilateral triangle is

$3a$ say

$LD$
In

$\displaystyle \Delta LMD$ , we have

$(LM)^{2}=(LD)^{2}+(MD)^{2}$

$\Rightarrow \displaystyle (LM)^{2}=9a^{2}+\left(\frac{LM}{2}\right)^{2}$

$\displaystyle\Rightarrow \frac{3}{4}(LM)^{2}=9a^{2}$

$\Rightarrow (LM)^{2}=12a^{2}$

Again in triangle

$\displaystyle OMD$ ,

$(OM)^{2}=(OD)^{2}+(MD)^{2}$

$\Rightarrow \displaystyle R^{2}=(3a-R)^{2}+ \left(\frac{LM}{2}\right)^{2}$

$\displaystyle\Rightarrow R^{2} =9a^{2}+R^{2}-6aR+3a^{2}$

$\displaystyle \Rightarrow 6aR=12a^{2}$

$\Rightarrow R=2a$
So, equation of circle is

$\displaystyle (x-0)^{2}+(y-0)^{2}=(2a)^{2}$

$\Rightarrow x^{2}+y^{2}=4a^{2}$

If two vertices of an equilateral triangle are

$(-1,0)$ and

$(1, 0),$ the equation of its circumcircle is

0%
$\sqrt{3} x^{2} +\sqrt{3}y^{2} +2y -\sqrt{3} =0$
0%
$2x^{2}+2y^{2}+\sqrt{3}y-2=0$
0%
$\sqrt{3} x^{2} +\sqrt{3}y^{2} -2y -\sqrt{3} =0$
0%
$2x^{2}+2y^{2}-\sqrt{3}y-2=0$

Explanation

Since two of the vertices

$A(-1,0)$ and

$B(1,0)$ lie on x-axis,

the third vertex

$C$ lies on y-axis (figure)

Let the coordinate of

$C$ be

$(0, x)$
Then

$(CA)^{2} =(AB)^{2} =(BC)^{2} =4.$

$\Rightarrow 1+ x^{2} = 4\Rightarrow x^{2}=3\Rightarrow x=\pm \sqrt{3} \Rightarrow OC=\pm \sqrt{3}$
Since triangle is equilateral, circurncentre of the triangle coincides with its centroid,

i.e. with

$(0,\pm 1\sqrt{3})$ and circurnradius is

$\dfrac 2{\sqrt{3}}$ .

Hence possible equations of the circumcircle is

$\displaystyle x^{2}\left( y\pm \frac{1}{\sqrt{3}}\right)^{2}=\left( \frac{2}{\sqrt{3}}\right)^{2}$

$\Rightarrow\displaystyle x^{2}+y^{2} \pm \frac{2}{\sqrt{3}} y-1=0$

$\displaystyle\Rightarrow \sqrt{3}x^{2}+\sqrt {3} y^{2} \pm 2y-\sqrt{3} =0$

Find the latus rectum of the parabola

$x^2\, +\, 2y- 3x\, +\, 5\, =\, 0$

0%
$1$
0%
$2$
0%
$4$
0%
$8$

Explanation

$x^2\, +\, 2y- 3x\, +\, 5\, =\, 0$

$\Rightarrow x^2-3x=-2y-5$

$\displaystyle \Rightarrow x^2-2.\frac{3}{2}x+\frac{9}{4}=-2y-5+\frac{9}{4}=-2y-\frac{11}{4}$

$\Rightarrow \displaystyle \left(x-\frac{3}{2}\right)^2=-2\left(y-\frac{11}{4}\right)$
Hence latus rectum

$=4\times \cfrac{1}{2}=2$

The lines

$2x-3y=5$ and

$3x-4y=7$ are diameters of a circle of area

$154\ sq.\ units$ . The equation of the circle is-

0%
${ x }^{ 2 }+{ y }^{ 2 }-2x-2y=47$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x-2y=62$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=47$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=62$

Explanation

The centre of circle is the point of intersection of the given diameter

$2x-3y=5$ and

$3x-4y=7$
which is

$\left( 1,-1 \right)$ and if r is the radius of the circle ,then

$\displaystyle { \pi }r^{ 2 }=154\Rightarrow { r }^{ 2 }=154\times \frac { 7 }{ 22 } \Rightarrow r=7$
Hence an equation of the required circle is

${ \left( x-1 \right) }^{ 2 }+{ \left( y+1 \right) }^{ 2 }={ 7 }^{ 2 }\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-2x+2y=47$

Find the equation of the circle whose centre is the point of intersection of the lines

$2x-3y+4=0$ and

$3x+4y-5=0$ and passes through the origin.

0%
$17({ x }^{ 2 }+{ y }^{ 2 })+2x-44y=0$
0%
$17({ x }^{ 2 }-{ y }^{ 2 })+2x+44y=0$
0%
$17({ x }^{ 2 }+{ y }^{ 2 })-2x+44y=0$
0%
None of these

Explanation

The point of intersection of the lines

$2x-3y+4=0$ and

$3x+4y-5=0$ is

$\left(\displaystyle-\frac{1}{17},\frac{22}{17}\right)$
Equation of circle whose centre is

$\left(\displaystyle-\frac{1}{17},\frac{22}{17}\right)$ and passing through origin is

$x^2+y^2+\displaystyle\frac{2x}{17}-\displaystyle\frac{44y}{17}=0$

$\Rightarrow 17(x^2+y^2)+2x-44y=0$
Hence, option A.

If two distinct chords of a parabola

$x^2\, =\, 4ay$ passing through

$(2a, a)$ are bisected on the line

$x + y = 1$ , then length of latus rectum can be

0%
$2$
0%
$1$
0%
$4$
0%
$5$

Explanation

Any point on the line $x+ y= 1$ can betaken as $(t,1-t)$

Equation of chord with this as midpoint is

${y(1-t)-2a(x+t) = (1-t)^2-4at}$

It passes through $(a,2a)$

therefore, ${t^2-2t+2a^2-2a+1=0}$

This should have two distinct real roots so

$-a<0$

$0<a<1$

length of latus rectum $< 4$

so answer is $1,2$

The equation of the circle passing through

$(3,6)$ and whose centre is

$(2,-1)$ is-

0%
${ x }^{ 2 }+{ y }^{ 2 }-4x+2y=45$
0%
${ x }^{ 2 }+{ y }^{ 2 }-4x-2y+45=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+4x-2y=45$
0%
${ x }^{ 2 }+{ y }^{ 2 }-4x+2y+45=0$

Explanation

$\textbf{Step -1: Find the radius of the circle.}$

$\text{As we know that, the standard equation of a circle is:}$

$(x-h)^2+(y-k)^2=r^2\text{ where }(h,k)\text{ is the centre and }r\text{ is the radius.}$

$\text{Here, the equation of the circle will be,}$

$(x-2)^2+(y+1)^2=r^2\ldots(i)$

$\text{and the point }(3,6)\text{ will satisfy this equation.}$

$\therefore (3-2)^2+(6+1)^2=r^2$

$\Rightarrow 1+49=r^2$

$\Rightarrow r^2=50$

$\textbf{Step -2: Find the equation of the required circle.}$

$\text{On putting the value of }r^2\text{ in equation }(i),\text{ we get}$

$(x-2)^2+(y+1)^2=50$

$\Rightarrow x^2+4-4x+y^2+1+2y=50$

$\Rightarrow x^2+y^2-4x+2y=45$

$\textbf{Hence, option A is correct.}$

The equation of the circle drawn with the focus of the parabola

$(x-1)^2 - 8y = 0$ as its centre and touching the parabola at its vertex is

0%
$x^2+y^2 - 4y = 0$
0%
$x^2+y^2 - 4y + 1=0$
0%
$x^2+y^2 - 2x - 4y = 0$
0%
$x^2 + y^2 - 2x - 4y + 1=0$

Explanation

$(x-1)^{2}=8y$

REF.Image

$a=2$

focus =

$(1,2)$ center of circle

Radius of circle =

$2$

$\Rightarrow$ Eq

$^n$

$\Rightarrow (x-1)^{2}+(y-2)^{2}=4$

$\Rightarrow \boxed {x^{2}-2x+y^{2}-4y+1=0}$ Ans

D is correct option.

1084757_262576_ans_0e210a561b944ebf920c726e2b00ce14.png

The intercept on the line

$y=x$ by the circle

${ x }^{ 2 }+{ y }^{ 2 }-2x=0$ is

$AB$ . Equation of the circle with

$AB$ as a diameter is

0%
${ x }^{ 2 }+{ y }^{ 2 }+x+y=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-x-y=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+x-y=0$
0%
None of these

Explanation

Equation of any circle passing through the point of intersection of

${ x }^{ 2 }+{ y }^{ 2 }-2x=0$ and

$y=x$ is

${ x }^{ 2 }+{ y }^{ 2 }-2x+\lambda \left( y-x \right) =0$

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\left( 2+\lambda \right) x=\lambda y=0.$

Its centre is

$\displaystyle \left( \frac { 2+\lambda }{ 2 } ,\frac { -\lambda }{ 2 } \right) .$

For

$AB$ to be the diameter of the required circle, the centre must lie on

$AB,$

i.e.,

$\displaystyle \frac { 2+\lambda }{ 2 } =\frac { -\lambda }{ 2 } \Rightarrow \lambda =-1.$

Thus, equation of required circle is

${ x }^{ 2 }+{ y }^{ 2 }-x-y=0.$

The normal at the point

$(3, 4)$ on a circle cuts the circle again at the point

$(1, 2)$ . Then the equation of the circle is -

0%
$\displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0$
0%
$\displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x+6y-12=0$
0%
$\displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x-6y+11=0$
0%
$\displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x+6y+11=0$

Explanation

We know the normal of a circle passes through it's center.
So the given points

$(3,4)$ and

$(1,2)$ are end point of the diameter.
and the centre of the circle is mid point of the given points,

$\displaystyle \equiv \left( 2,3 \right)$
and Radius

$=\displaystyle \frac { \sqrt { 4+4 } }{ 2 } =\frac { 2\sqrt{ 2 } }{ 2 } =\sqrt { 2 }$
Hence required circle is,

$\displaystyle (x-2{ ) }^{ 2 }+(y-3{ ) }^{ 2 }=2$

$\Rightarrow \displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x-6y+11=0$

The lines

$2x - 3y = 5 \ \ \& \ \ 3x - 4y = 7$ are diameters of a circle of area 154 sq units. Then the equation of the circle is

0%
$\displaystyle x^{2}+y^{2}+2x-2y=62$
0%
$\displaystyle x^{2}+y^{2}-2x+2y=47$
0%
$\displaystyle x^{2}+y^{2}+2x-2y=47$
0%
$\displaystyle x^{2}+y^{2}-2x+2y=62$

Explanation

Intersection of any 2 diameters gives us the center

$2x – 3y = 5 -eq.1$

$3x – 4y = 7 -eq.2$

$3 \times eq.1 – 2 \times eq.2$

$\implies -9y + 8y = 15 - 14$

$\implies y = -1 \implies x = 1$

Center $= (1,-1)$

And given $A = 154$

$\implies \pi r^2 = 154$

$r^2 = \dfrac{154}{22} \times 7 = 49$

$r = 7$

Equation of circle is

$(x - 1) ^2 + (y+1)^2 = 7^2$

$\implies x^2 + y^2 -2x + 2y = 47$

For the points on the circle

$\displaystyle x^{2}+y^{2}-2x-2y+1=0$ , the sum of maximum and minimum values of

$4x + 3y$ is

0%
$\displaystyle \frac{26}{3}$
0%
$10$
0%
$12$
0%
$14$

Explanation

The given equation can be written as $(x - 1) ^2 + (y - 1)^2 = 1^2$

Any point on this circle is given by $(1 + \cos \theta , 1 + \sin \theta)$

$4x + 3y = 7 + 4 \cos \theta + 3 \sin \theta = f(\theta)$

If $f = a \cos \theta + b \sin \theta + c$

$max = c + \sqrt{a^2 + b^2} , min = c - \sqrt{a^2 + b^2}$

$\implies max \, f(\theta) + min \, f(\theta) = 2c =2 \times 7 = 14$

The lines

$2x - 3y = 5$ &

$3x - 4y = 7$ are diameters of a circle of area

$154$ sq units Then the equation of the circle is

0%
$\displaystyle x^{2}+y^{2}+2x-2y=62$
0%
$\displaystyle x^{2}+y^{2}-2x+2y=47$
0%
$\displaystyle x^{2}+y^{2}+2x-2y=47$
0%
$\displaystyle x^{2}+y^{2}-2x+2y=62$

Explanation

Intersection of any 2 diameters gives us the center

$2x – 3y = 5 -eq.1$

$3x – 4y = 7 -eq.2$

$3 \times eq.1 – 2 \times eq.2$

$\implies -9y + 8y = 15 - 14$

$\implies y = -1 \implies x = 1$

Center = (1,-1)

And given A = 154

$\implies \pi r^2 = 154$

$r^2 = \dfrac{154}{22} \times 7 = 49$

$r = 7$

Equation of circle is

$(x - 1) ^2 + (y+1)^2 = 7^2$

$\implies x^2 + y^2 -2x + 2y – 47 = 0$

The centre of a circle is

$( x -2 , x+1 )$ and it passes through the points

$( 4 , 4 )$ Find the value ( or values ) of

$x$ , if the diameter of the circle is of length

$\displaystyle 2\sqrt{5}$ units.

0%
$1$ or $3$
0%
$-1$ or $4$
0%
$5$ or $4$
0%
$3$ or $-2$

Explanation

Radius of the circle

$=$ dist. between the center and given pt. on the circle.

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula

$\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Distance between the points $(x-2,x+1)$ and D $(4,4)$

$= \sqrt { \left( 4-x + 2 \right) ^{ 2 }+\left( 4 - x - 1 \right) ^{ 2 } }$

$= \sqrt { \left( 6-x \right) ^{ 2 }+\left( 3 - x \right) ^{ 2 } }$

$= \sqrt { 36 + {x}^{2} - 12x + 9 + {x}^{2} - 6x } = \sqrt { 2{x}^{2} -18x + 45 }$

Given, diameter $= 2 \sqrt {5}$ $\Rightarrow$ Radius $= \sqrt {5}$

$\Rightarrow \sqrt { 2{x}^{2} -18x + 45 } = \sqrt {5}$
Squaring both sides,

$2{x}^{2} -18x + 45 = 5$

$2{x}^{2} -18x + 40 = 0$

${x}^{2} -9x + 20 = 0$

$(x-5)(x-4) = 0$

$x = 4$ or

$5$

The equation circle whose center is

$(0,0)$ and radius is

$4$ is

0%
$x^2+y^2=4$
0%
$x^2+y^2=16$
0%
$x^2+y^2=2$
0%
None.

Explanation

The equation of circle is

$x^2+y^2=r^2$

Here the radius is

$4$

So the equation is

$x^2+y^2=4^2\\x^2+y^2=16$

The equation

$\displaystyle \frac {x^2}{8-t}\, +\, \displaystyle \frac {y^2}{t-4}\, =\, 1$ will represent an ellipse if

0%
$t\, \in\, (1,\, 5)$
0%
$t\, \in\, (2,\, 8)$
0%
$t\, \in\, (4,\, 8)\, -\, \{6\}$
0%
$t\, \in\, (4,\, 10)\, -\, \{6\}$

Explanation

Consider Equation,

$\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$ to represent an ellipse equation.

$a>0,b>0,a\neq b$
Given,equation

$\displaystyle\frac{x^2}{(8-t)}+\displaystyle\frac{y^2}{(t-4)}=1$

$\Rightarrow (8-t)>0\;$ and

$\;(t-4)>0,(8-t)\neq(t-4)$

$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$ {

$t\neq6$ }

$\Rightarrow t\in(4,8)-$ {

$6$ }

Two circles touch each other externally at C and a common tangent touches them at A and B. Which one is true?

0%
$CD\perp AB$
0%
$\angle ACB=90^o$
0%
$AC=BC$
0%
All of these

Explanation

According to the question

Lets suppose that,

X and Y are two circle touch each other at P.

AB is the common tangent to circle X and Y at point A and B.

According In the given figer,

In triangle $PAC,\ \angle CAP=\angle APC=\alpha$

Similarly $CB=CP,\ \angle CPB=\angle PBC=\beta$

Now triangle APB,

$\angle PAB+\angle PBA+\angle APB=180$

$\alpha +\beta +\left( \alpha +\beta \right)=180$

$2\alpha +2\beta =180$

$\alpha +\beta =90$

$\therefore \ \angle APB=90=\alpha +\beta .$

This is the required solution.

1177630_379399_ans_733db3414c204b2eb882ac74be8def7d.png

Find the center-radius form of the equation of the circle with center

$\left( 4,0 \right)$ and radius

$7$

0%
${ \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }=49$
0%
${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=7$
0%
${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=7$
0%
${ \left( x+4 \right) }^{ 2 }+{ y }^{ 2 }=49$

Explanation

If $(-g,-f)$ is the center and $r$ is radius

The $(x + g) ^2 + (y+ f)^2 = r^2$ is the equation of the circle

There $C = (4,0) , r = 7$

$\implies (x - 4) ^2 + (y - 0)^2 = 7^2$

$(x – 4) ^2 + (y)^2 = 49$

What is the equation of a circle with center (-3,1) and radius 7?

0%
$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=7$
0%
$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=49$
0%
$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=7$
0%
$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=49$

Explanation

the general equation of a circle with center at (a,b) and radius r is

$(x-a)^2+(y-b)^2=r^2$

so substituting the values we get

the equation of the circle is

$(x+3)^2+(y-1)^2=7^2=49$

An ambulance company provides services within an 80 mile radius of their headquarters If this service area is represented graphically with the headquarters located at the coordinates (0, 0) what is the equation that represents the service area?

0%
$\displaystyle x^{2}+y^{2}=80$
0%
$\displaystyle \left ( x-0 \right )^{2}+\left ( y-0 \right )^{2}=80$
0%
$\displaystyle x^{2}+y^{2}=1600$
0%
$\displaystyle x^{2}+y^{2}=6400$

Explanation

the general equation of a circle with center at (a,b) and radius r is

$(x-a)^2+(y-b)^2=r^2$

so substituting the values we get the circle equation as

$x^2+y^2=80^2=6400$ option D

Graph the circle

${ \left( x-3 \right) }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=4$

Explanation

$(x - 3) ^2 + (y+ 4)^2 = 4$

$\implies center = (3,-4) , r = 2$

Since $C = (3,-4)$

C lies in the fourth quadrant with radius

$2$

The point

$P(9/2, 6)$ lies on the parabola

$y^2=4ax$ , then parameter of the point P is

0%
$\frac{3a}{2}$
0%
$\frac{2}{3a}$
0%
$\frac{2}{3}$
0%
$\frac{3}{2}$

Explanation

As point

$P\left( \dfrac { 9 }{ 2 } ,6 \right)$ lies on the parabola

${ y }^{ 2 }=4ax$ , it must satisfy the equation of parabola.

$\therefore { 6 }^{ 2 }=4a\left( \dfrac { 9 }{ 2 } \right)$

$\therefore 36=18a$

$\therefore a=2$

Now, point P which lies on parabola has coordinates

$P\left( a{ t }^{ 2 },2at \right)$

Where, t is a parameter

$\therefore a{ t }^{ 2 }=\dfrac { 9 }{ 2 }$

$\therefore 2{ t }^{ 2 }=\dfrac { 9 }{ 2 }$

$\therefore { t }^{ 2 }=\dfrac { 9 }{ 4 }$

$\therefore { t }=\dfrac { 3 }{ 2 }$

Find the equation of the circle with center on x + y = 4 and 5x + 2y + 1 = 0 and having a radius of 3

0%
$\displaystyle x^{2}+y^{2}+6x-16y+64=0$
0%
$\displaystyle x^{2}+y^{2}+8x-14y+25=0$
0%
$\displaystyle x^{2}+y^{2}+6x-14y+49=0$
0%
$\displaystyle x^{2}+y^{2}+6x-14y+36=0$
0%
none of these

Explanation

Given,

Centre of circle is on

$x+y=4$ and

$5x+2y+1=0$

Thus,

$x+y=4$

$=>y=4-x$ (i)

$5x+2y+1=0$ (ii)

$=>5x+2(4-x)+1=0$

$=>5x+8-2x+1=0$

$=>3x+9=0$

$=>x=-3$

$\therefore y=4-(-3)$

$=7$

$\therefore$ Centre of circle is

$(-3,7)$

Radius of circle=

$3$ (given)

$\therefore$ Equation of circle is

$(x-(-3))^2+(y-7)^2=3^2$

$=>(x+3)^2+(y-7)^2=9$

$=>(x^2+6x+9)+(y^2-14y+49)-9=0$

$=>x^2+6x+y^2-14y+49=0$

Find the equation of k for which the equation

$\displaystyle x^{2}+y^{2}+4x-2y-k=0$ represents a point circle

0%
$5$
0%
$-5$
0%
$6$
0%
$-6$
0%
none of these

Explanation

Standard form of circle is

$(x-x_1)^2+(y-y_1)^2=r^2$

When

$(x_1,y_1)$ is the centre of the circle and

$r$ is the radius of the circle.

Now,given equation is,

$x^2+y^2+4x-2y-k=0$

$=>x^2+4x+y^2-2y=k$

$=>x^2+2.x.2+2^2+y^2-2.y.1+1^1=k+2^2+1^1$

$=>(x+2)^2+(y-1)^2=k+4+1$

$=>(x+2)^2+(y-1)^2=k+5$

Since it represents a point circle

$\therefore k+5=0$

$=>k=-5$

The equation

$y^{2} + 4x + 4y + k = 0$ represents a parabola whose latus rectum is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$y^{2} + 4x + 4y + k = 0$

$y^{2} + 2\times 2y + 4 - 4 + 4x + k = 0$

$(y + 2)^{2} = -4x - k + 4$

$(y + 2)^{2} = -4 \left (x - \dfrac {4 + k}{4}\right )$

$\therefore$ Latus rectum

$= 4$ units

Equation of the parabola with its vertex at

$(1, 1)$ and focus

$(3, 1)$ is

0%
$(x 1)^2 = 8 (y - 1)$
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$(y - 1)^2 = 8 (x - 3)$
0%
$(y - 1)^2 = 8 (x - 1)$
0%
$(x - 3)^2 = 8(y - 1)$

Explanation

As the vertex is $(1,1)$ and focus is $(3,1)$ , whose ordinate is same its axis of symmetry is $y = 1$ .

And as vertex is equidistant from foci and directrix, and latter is perpendicular to axis of symmetry.

Directrix is $x =1$

As parabola is the locus of a point whose distance from directrix $x+1 = 0$ and focus $(3,1)$

Its equation is $(x-3)^2+(y-1)^2= (x+1)^2$

$\Rightarrow x^2-6X+9+y^2-2Y+1 = x^2+2X+1$

$\Rightarrow y^2-2y+9 = 8x$

$\Rightarrow (y-1)^2 = 8(x-1)$

So, option C is the answer.

The equation of a circle which has a tangent

$3x+4y=6$ and two normals given by

$(x-1)(y-2)=0$ is

0%
$(x-3)^2+(y-4)^2=5^2$
0%
$x^2+y^2-4x-2y+4=0$
0%
$x^2+y^2-2x-4y+4=0$
0%
$x^2+y^2-2x-4y+5=0$

Explanation

Equation of tangent

$=3x+4y=6$ and two normals are

$(x-1)(y-2)=0$

$\Rightarrow x-1=0\,$ and

$\, y-2=0$

$\therefore \text{Radius}=\displaystyle \frac{3(1)+4(2)-6}{\sqrt{9+16}}$

$=\displaystyle \frac{5}{5}=1$

$\therefore$ Equation of the circle is

$(x-1)^2+(y-2)^2=1\,$ or

$\, x^2+y^2-2x-4y+4=0$

The ends of the latus rectum of the conic

$x^{2} + 10x - 16y + 25 = 0$ are

0%
$(3, -4), (13, 4)$
0%
$(-3, -4), (13, -4)$
0%
$(3, 4), (-13, 4)$
0%
$(5, -8), (-5, 8)$

Explanation

Given that

$x^{2} + 10x - 16y + 25 = 0$

$\Rightarrow (x + 5)^{2} = 16y$

$\Rightarrow X^{2} = 4AY$ ,
where

$X = x + 5, A = 4, Y = y$
The ends of the latus rectum are

$(2A, A)$ and

$(-2A, A)$

$\Rightarrow x + 5 = 2(4)$

$\Rightarrow x = -8 - 5 = 3, y = 4$
and

$x + 5 = -2(4)$

$\Rightarrow x = -8 - 5 = -13, y = 4$

$\Rightarrow (3, 4)$ and

$(-13, 4)$ .

The equation of the circle whose centre and radius are

$\left( 1,-1 \right)$ and

$4$ respectively, is

0%
${ x }^{ 2 }-{ y }^{ 2 }+2x+2y-14=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-14=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+14=0$

Explanation

We know that equation of circle is

${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }$
Here, centre is

$\left( 1,-1 \right)$ and radius is

$4$ .

$\therefore$ Equation is

${ \left( x-1 \right) }^{ 2 }+{ \left( y+1 \right) }^{ 2 }={ \left( 4 \right) }^{ 2 }$

$\Rightarrow { x }^{ 2 }+1-2x+{ y }^{ 2 }+1+2y=16$

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$

For hyperbola

$\dfrac{x^2}{cos^2a}-\dfrac{y^2}{sin^2a}=1$ which of the following remains constant with change in 'a'?

0%
Abscissae of vertices
0%
Abscissae of foci
0%
Eccentricity
0%
Directrix

Explanation

Comparing given equation

$\dfrac{x^2}{cos^2a}-\dfrac{y^2}{sin^2a}=1$ with the standard equation

$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} =1$ , we get

$a^2 = cos^2 a$ and

$b^2 = sin^2 a$

$\therefore 1=sin^2 a + cos^2 a = a^2 + b^2$

$\Rightarrow e=\sqrt { \dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 } } }$

$\Rightarrow e=\dfrac{1}{cos\,a}$
Now, foci

$ae=cos\,a.\dfrac{1}{cos\,a}=1$

The equation of the ellipse having vertices at

$\displaystyle \left( \pm 5,0 \right)$ and foci

$\displaystyle \left( \pm 4,0 \right)$ is

0%
$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1$
0%
$\displaystyle 9{ x }^{ 2 }+25{ y }^{ 2 }=225$
0%
$\displaystyle \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1$
0%
$\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$

Explanation

The vertices and foci of an ellipse are

$\displaystyle \left( \pm 5,0 \right)$ and

$\displaystyle \left( \pm 4,0 \right)$ respectively.

$\displaystyle \therefore \quad a=5$ and

$\displaystyle ae=4$

$\displaystyle \Rightarrow \quad e=\frac { 4 }{ 5 }$
We know that,

$\displaystyle e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } }$

$\displaystyle \Rightarrow \quad \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 25 } \Rightarrow { b }^{ 2 }=9$
Hence, equation of an ellipse is

$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$

The equation of a parabola which passes through the intersection of a straight line x

$+$ y

$=0$ and the circle

$x^2+y^2+4y=0$ is.

0%
$y^2=4x$
0%
$y^2=x$
0%
$y^2=2x$
0%
None of these

Explanation

$x+y=0$

$x^2+y^2+4y=0$

The line and the circle intersect at the point

$(0,0)$ and

$(2,-2)$

Among the given options, only

$y^2=2x$ passes through both these points.

So, the answer is option (C).

The distance between the foci of the hyperbola

${ x }^{ 2 }-3{ y }^{ 2 }-4x-6y-11=0$ is

0%
$4$
0%
$6$
0%
$8$
0%
$10$

Explanation

Given, equation of hyperbola is

${ x }^{ 2 }-3{ y }^{ 2 }-4x-6y-11=0$

$\Rightarrow \left( { x }^{ 2 }-4x+4 \right) -3\left( { y }^{ 2 }+2y+1 \right) -11=4-3$

$\Rightarrow { \left( x-2 \right) }^{ 2 }-3{ \left( y+1 \right) }^{ 2 }=12$

$\Rightarrow \dfrac { { \left( x-2 \right) }^{ 2 } }{ 12 } -\dfrac { { \left( y+1 \right) }^{ 2 } }{ 4 } =1$
Now,

$e=\sqrt { 1+\dfrac { 4 }{ 12 } } =\dfrac { 2 }{ \sqrt { 3 } }$

$\therefore$ Distance between foci

$=2ae=2\times \sqrt { 12 } \times \dfrac { 2 }{ \sqrt { 3 } } =8$

The equation to the circle with centre

$(2, 1)$ and touching the line

$3x + 4y = 5$ is

0%
$x^{2} + y^{2} - 4x - 2y + 5 = 0$
0%
$x^{2} + y^{2} - 4x - 2y - 5 = 0$
0%
$x^{2} + y^{2} - 4x - 2y + 4 = 0$
0%
$x^{2} + y^{2} - 4x - 2y - 4 = 0$

Explanation

The distance from the line

$ax+by+c=0$ from the line to a point

$(x_0,y_0)$ is:

$D=\dfrac { \left| a{ x }_{ 0 }+b{ y }_{ 0 }+c \right| }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } }$

Here, the distance is equal to the radius of the circle and the distance from the line

$3x+4y-5=0$ to a point

$(2,1)$ is:

$r=\dfrac { \left| (3\times 2)+(4\times 1)+(-5) \right| }{ \sqrt { 3^{ 2 }+{ 4 }^{ 2 } } } =\dfrac { \left| 6+4-5 \right| }{ \sqrt { 9+16 } } =\dfrac { \left| 5 \right| }{ \sqrt { 25 } } =\dfrac { 5 }{ 5 } =1$

Therefore, equation of the circle with centre

$(2,1)$ and radius

$1$ is:

${ \left( x-2 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ \left( 1 \right) }^{ 2 }\\ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+1-2y=1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-2y+5=1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-2y+4=0$

Hence, the equation of circle is

$x^2+y^2-4x-2y+4=0$ .

Find the equation of a circle with center

$(2,0)$ and passing through point

$\left( 3,\sqrt { 3 } \right)$ .

0%
$(x - 2)^{2} + y^{2} = 4$
0%
$(x + 2)^{2} + y^{2} = 4$
0%
$(x - 2)^{2} + y^{2} = 16$
0%
$(x + 2)^{2} + y^{2} = 16$

Explanation

Let the radius be '

$r$ '.

Then the equation of the circle will be

$(x-2)^{2}+y^{2}=r^{2}$

Now, substituting the point

$(3,\sqrt{3})$ in the above equation gives us

$(3-2)^{2}+(\sqrt{3})^{2}=r^{2}$

$\Rightarrow 1+3^{2}=r^{2}$ or

$r^{2}=4$

Hence, the equation of the circle is

$(x-2)^{2}+y^{2}=4$ .

$(a,b)$ lies on circle with centre as origin, then its radius will be

0%
$\displaystyle a-b$
0%
$\displaystyle a+b$
0%
$\displaystyle \sqrt { { a }^{ 2 }+{ b }^{ 2 } }$
0%
$\displaystyle { a }^{ 2 }+{ b }^{ 2 }$

Explanation

We know the formula,

The equation of a circle of radius r and centre the origin is

$x^2+y^2=r^2$

Here the center is (a, b)

so Radius , $r = \sqrt{a^2+b^2}$

The radius of the circle

$x^{2} + y^{2} + 4x + 6y + 13 = 0$ is

0%
$\sqrt {26}$
0%
$\sqrt {13}$
0%
$\sqrt {23}$
0%
$0$

Explanation

Given equation is

$x^{2} + y^{2} + 4x + 6y + 13 = 0$
or

$(x^{2} + 4x + 4) + (y^{2} + 6y + 9) + 13 = 4 + 9$
or

$(x + 2)^{2} (y + 3)^{2} = 0$

$\therefore$ Radius of circle

$= 0$ .

The sum of the focal distances of any point on the conic

$\dfrac {x^{2}}{25} + \dfrac {y^{2}}{16} = 1$ is

0%
$10$
0%
$9$
0%
$41$
0%
$18$

Explanation

We know, if P is any point on the curve, then Sum of focal distances

$=$ length of major axis
i.e.,

$SP + S'P = 2a$

$= 2(5) [\because a^{2} = 5^{2}]$

$= 10$

The point

$(3, 4)$ is the focus and

$2x - 3y + 5 = 0$ is the directrix of a parabola. Lenghth of latus rectum is

0%
$\dfrac {2}{\sqrt {13}}$
0%
$\dfrac {4}{\sqrt {13}}$
0%
$\dfrac {1}{\sqrt {13}}$
0%
$\dfrac {3}{\sqrt {13}}$

Explanation

We know that, Latus rectum

$= 2\times$ (distance from focus to directrix)

$= 2\cdot \dfrac {|6 - 12 + 5|}{\sqrt {4 + 9}} = \dfrac {2}{\sqrt {13}}$

If the centre

$O$ of circle is the intersection of

$x-$ axis and line

$y=\dfrac { 4 }{ 3 } x+4$ , and the point

$(3,8)$ lies on circle, then the equation of circle will be

0%
$\displaystyle { x }^{ 2 }+{ y }^{ 2 }=25$
0%
$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=25$
0%
$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=100$
0%
$\displaystyle { \left( x+3 \right) }^{ 2 }+{ \left( y-8 \right) }^{ 2 }=100$

Explanation

The intersection of

$x$ -axis and the line

$y = \cfrac{4x}{3} + 4$ is

$(-3,0)$ , which is the centre of the circle.

Since the point

$(3,8)$ lies on the circle, its radius becomes the distance between these points, which is

$\sqrt{(3 + 3)^2 + (8 - 0)^2} = 10$

The equation thus becomes

$(x + 3)^2 + (y - 0)^2 = (10)^2$

i.e.

$(x + 3)^2 + y^2 = 100$

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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