MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 1

If the lines

$2\mathrm{x}+3\mathrm{y}+1=0$ and

$\mathrm{3x}- \mathrm{y}-4=0$ lie along diameters of a circle of circumference

$10\pi$ , then the equation of the circle is:

0%
$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}+2\mathrm{y}-23=0$
0%
$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}-2\mathrm{y}-23=0$
0%
$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}+2\mathrm{y}-23=0$
0%
$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}-2\mathrm{y}-23=0$

Explanation

Two diameters are along

$2x+3y+1=0$

$3x−y−4=0$

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

$(1)$

$3x−y−4=0$ ...............

$(2) \times 3$

Subtract

$(2)-(1)$

$\Rightarrow x =1$ and

$y=-1$

on solving we get center as

$(1,−1)$

We know that circumference

$=2\pi r$

$2\pi r=10\pi$

$2r=10$

$r=5$

Required circle is

$(x−1)^2+(y+1)^2=52$

$x^2+y^2−2x+2y−23=0$

The circle

$x^{2}+y^{2}-8x=0$ and hyperbola

$\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$ intersect at the points

$A$ and

$B$ .

then the equation of the circle with

$AB$ as its diameter is

0%
$x^{2}+y^{2}-12x+24=0$
0%
$x^{2}+y^{2}+12x+24=0$
0%
$x^{2}+y^{2}+24x-12=0$
0%
$x^{2}+y^{2}-24x-12=0$

Explanation

Given:

$x^2+y^2-8x=0 .....(1)$

$\implies y^2=8x-x^2$

$\dfrac {x^2}{9}-\dfrac {y^2}{4}\Rightarrow 1$

$\Rightarrow\dfrac {x^2}{9}-\dfrac {(8x-x^2)}{4}=1$

$\Rightarrow4x^2-9(8x-x^2)=36$

$\Rightarrow4x^2-72x+9x^2-36=0$

$\Rightarrow13x^2-72x-36=0$

$\Rightarrow x=\dfrac {72\pm \sqrt {(72)^2-4\times 13\times 36}}{26}$

$\Rightarrow x=6$

Then, putting

$x=6$ in equation (1), we get

$\Rightarrow y=\pm \sqrt{12}=\pm \sqrt{4\times3}=\pm 2\sqrt 3$

So, radius

$=2\sqrt 3$

$AB$ is a diameter center will be the midpoint of point

$A=(6,2\sqrt 3)$ and

$B=(6,-2\sqrt 3)$

By mid point formula,

Center

$=\left(\dfrac{6+6}{2},\dfrac{2\sqrt 3-2\sqrt 3}{2}\right)$

Centre

$=(6, 0)$

$\Rightarrow(x-6)^2+y^2=12$

$\Rightarrow x^2+36-12x+y^2=12$

$\Rightarrow x^2-12x+y^2+24=0$

The length of the latus rectum of the parabola

$169 \left[(x-1)^2+(y-3)^2\right]=(5x-12y+17)^2$ is:

0%
$\displaystyle \frac { 14 }{ 13 }$
0%
$\displaystyle \frac { 12 }{ 13 }$
0%
$\displaystyle \frac { 28 }{ 13 }$
0%
$none\ of\ these$

Explanation

Given equation can be rewritten as

$\displaystyle { \left( x-1 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }={ \left( \frac { 5x-12y+17 }{ 13 } \right) }^{ 2 }$

$\displaystyle \Rightarrow \quad SP=PM$
Here, focus is

$(1, 3)$ , directrix

$\displaystyle 5x-12y+17=0$

$\displaystyle \therefore$ the distance of the focus from the directrix

$\displaystyle =\left| \frac { 5-36+17 }{ \sqrt { 25+144 } } \right|$

$\displaystyle =\frac { 14 }{ 13 } =2a$

$\displaystyle \therefore$ Latusrectum

$\displaystyle =2\times \frac { 14 }{ 13 } =\frac { 28 }{ 13 }$

Equation of the ellipse in its standard form is

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

0%
True
0%
False
0%
Nither
0%
Either

Explanation

Equation of ellipse in standard form is

$\dfrac { { x }^{ 2 } }{ a^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$

False

The equation of the circle touching

$x = 0, y = 0$ and

$x = 4$ is

0%
$x^2 + y^2 - 4x - 4y + 16 = 0$
0%
$x^2 + y^2 - 8x - 8y + 16 = 0$
0%
$x^2 + y^2 + 4x + 4y + 4 = 0$
0%
$x^2 + y^2 - 4x - 4y + 4 = 0$

Explanation

The given circle is as shown in the following figure
It is clear that the coordinates of the centre of the circle are (2, 2) and radius = 2
Hence, the required equation of the circle is

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

$\Rightarrow x^2 + y^2 - 4x - 4y + 4 = 0$
1100017_357699_ans_6a7ccadc702f416eaeb0f4f7fb8c8f05.jpg

The radius of the circle with center (0,0) and which passes through (-6,8) is

0%
5
0%
10
0%
6
0%
8

Explanation

$\displaystyle r=\sqrt{\left ( 6 \right )^{2}+\left ( -8 \right )}=10$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

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

Radius

$=2$ Centre

$(-1,1)$

Assertion is incorrect but reason is correct.

The equation of circle with its centre at the origin is

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

0%
True
0%
False
0%
Neither
0%
Either

Explanation

The center is a fixed point in the middle of the circle; usually given the general coordinates (h, k).

The fixed distance from the center to any point on the circle is called the radius.

A circle can be represented by two different forms of equations, the general form and the center-radius form. This discussion will focus on the center-radius form.

centre radius form

let (h,k) be the centre of a cicle and r be the radius

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

and we talk about centre at origin

than $(h,k)$ will become $(0,0)$

so, equation will become ${x}^{2}+{y}^{2}={r}^{2}$

therefore option A will be answer.

Which of the following equations of a circle has center at (1, -3) and radius of 5?

0%
$\displaystyle x^{2}+y^{2}=25$
0%
$\displaystyle \left ( x-1 \right )^{2}+\left ( y+3 \right )^{2}=25$
0%
$\displaystyle \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=25$
0%
$\displaystyle \left ( x+1 \right )^{2}+\left ( y-3 \right )^{2}=25$

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

$(x-1)^2+(y+3)^2=25$ option B

Determine the area enclosed by the curve

$\displaystyle x^{2}-10x+4y+y^{2}=196$

0%
$15\pi$
0%
$225\pi$
0%
$20\pi$
0%
$17\pi$

Explanation

Equation of the circle is

$x^2-10x+4y+y^2=196$

$=>x^2-2\times x \times 5 +5^2+y^2+2\times y\times 2+2^2=196+2^2+5^2$

$=>(x-5)^2+(y+2)^2=196+4+25$

$=>(x-5)^2+(y+2)^2=225$

$=>(x-5)^2+(y+2)^2=15^2$

Thus radius of the circle is 15

Thus, area of the circle

$=\pi \times 15^2$

$=225\pi$

The standard equation of circle at origin is

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

Explanation

The standard equation of circle at origin is

$x^2+y^2=r^2$
Example: A circle with its center at the origin

$(0, 0)$ , and radius

$r$ has the equation

$x^2 + y^2 = r^2$ . We'll know if a given point is on the circle if the coordinates of that point satisfy the equation.

The diameter of a circle described by

$\displaystyle 9x^{2}+9y^{2}=16$ is

0%
$\dfrac {16}9$
0%
$\dfrac 43$
0%
$4$
0%
$\dfrac 83$

Explanation

Equation of circle is

$9x^2+9y^2=16$

$=>x^2+y^2=\dfrac{16}{9}$

$=>x^2+y^2=(\dfrac{4}{3})^2$

$\therefore$ Radius of the circle =

$\dfrac{4}{3}$

$\therefore$ Diameter of the circle=

$\dfrac{4\times 2}{3}$

$=\dfrac{8}{3}$

A circle has a diameter whose ends are at (-3, 2) and (12, -6) Its Equation is

0%
$\displaystyle 4x^{2}+4y^{2}-36x+16y+192=0$
0%
$\displaystyle 4x^{2}+4y^{2}-36x+16y-192=0$
0%
$\displaystyle 4x^{2}+4y^{2}-36x-16y-192=0$
0%
$\displaystyle 4x^{2}+4y^{2}-36x-16y+192=0$

Explanation

Coordinate of diameter is

$(-3,2)$ and

$(12,-6)$

Centre of the circle is the midpoint of the diameter.

$\therefore (x_1,y_1)=(\dfrac{-3+12}{2}, \dfrac{2-6}{2})$

$=(\dfrac{9}{2}, \dfrac{-4}{2})$

$=(\dfrac{9}{2}, -2)$

Half the distance between

$(-3,2)$ and

$(12,-6)$ is the radius.

$\therefore 2r=\sqrt {(-3-12)^2+(2+(-6))^2}$

$=>2r=\sqrt {15^2+8^2}$

$=>2r=\sqrt {225+64}$

$=>2r=17$

$=>r=\dfrac{17}{2}$

$$\therefore Equation of circle is

$x-(\dfrac{9}{2})^2+(y-(-2))^2=(\dfrac{17}{2})^2$

$=>\dfrac{(2x-9)^2}{4}+(y+2)^2=\dfrac{289}{4}$

$=>4x^2-36x+81+4(y^2+4y+4)-289=0$

$=>4x^2-36x+81+4y^2+16y+16-289=0$

$=>4x^2+4y^2-36x+16y+192=0$

What is the nature of the given graph?

0%
The graph in symmetric about x-axis
0%
The graph in symmetric about y-axis
0%
Sum of x-intercept and y-intercept is greater than zero
0%
The polynomial has $3$ terms

Explanation

R.E.F image

As graph is touching x axis

& graph is increasing on both

side about is axis

$\therefore$ has 2

equal roots at

$\boxed{x = 0};$ hence

$2^{\circ}$ Polynomial.

$\Rightarrow y = ax^{2}+bx+c$

$\Rightarrow \boxed{y = x^{2}}$ using

$\Rightarrow$

$\begin{bmatrix} At\,x = 0 ; y =0 \\ x=1;y=1 \\ x=-1;y=1 & \end{bmatrix}$

1104218_517088_ans_ed3a679d757746c5a0c00e38ab00f056.png

Find the equation of a circle with center

$(0, 0)$ and radius

$5$ .

0%
$x^2+y^2=5$
0%
$x^2-y^2=25$
0%
$x^2+y^2=25$
0%
$(x-1)^2+(y+1)^2=25$

Explanation

Compare the equation with the standard form with center at

$(h, k)$ and radius

$r$ is:

$(x-h)^2+(y-k)^2=r^2$
Substitute the value of

$(h, k) = (0, 0)$ and

$r = 5$
Then, the equation becomes

$x^2+y^2 = 25$

What is the radius of the circle with the following equation?

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

0%
$3.46$
0%
$5$
0%
$7$
0%
$6$

Explanation

Equation of circle is

$x^2-6x+y^2-4y-12=0$

$=>x^2-2.3.x+y^2-2.2y-12=0$

$=>x^2-2.3.x+3^2+y^2-2.2.y+2^2=12+3^2+2^2$

$=>(x-3)^2+(y-2)^2=25$

$=>(x-3)^2+(y-2)^2=5^2$

$\therefore$ Radius of the circle=

$5$

The locus of a planet orbiting around the sun is:

0%
A circle
0%
A straight line
0%
A semicircle
0%
An ellipse

Number of intersecting points of the conic

$4x^{2} + 9y^{2} = 1$ and

$4x^{2} + y^{2} = 4$ is

0%
$1$
0%
$2$
0%
$3$
0%
$0$ (zero)

Explanation

Number of intersecting points of the conic

Lets assume:

$C_1: 4x^{2} + 9y^{2} = 1$

$C_2: 4x^{2} + y^{2} = 4$

They can be written as:

$C_{1} : \dfrac {x^{2}}{\dfrac {1}{4}} + \dfrac {y^{2}}{\dfrac {1}{9}} = 1$

$C_{2} : \dfrac {x^{2}}{1} + \dfrac {y^{2}}{4} = 1$

Plotting these on coordinate axes, we get the following figure.

We can see that there is no intersection.

Hence, zero point of intersections.

The point (3,4) is the focus and

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

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

Explanation

$4a=2$ (distance from focus to directrix)

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

The circle with radius

$1$ and centre being foot of the perpendicular from

$(5, 4)$ on y-axis, is?

0%
$x^2+y^2-8x-15=0$
0%
$x^2+y^2-10x+24=0$
0%
$x^2+y^2-8y+15=0$
0%
$x^2+y^2+2y=0$

Explanation

Foot of perpendicular of (5,4) on y-axis is

$(0,4)$

Therefore the equation of circle with radius

$1cm$ is

${ (x-0) }^{ 2 }+{ (y-4) }^{ 2 }=1$

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-8y+16=1$

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-8y+15=0$

Equation of the circle with centre on y-axis and passing through the points

$(1,0),(1,1)$ is:

0%
${ x }^{ 2 }+{ y }^{ 2 }-y-1=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-x-1=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-x+1=0\quad$
0%
${ x }^{ 2 }+{ y }^{ 2 }-y+1=0$

Explanation

Let the centre of the circle be

$(0,k)$ .

Then, the equation of the circle is

$(x-0)^2+(y-k)^2=r^2$

Circle passes through the points

$(1,0)$ and

$(1,1)$ .

Thus,

$1+k^2=r^2....(1)$ and

$1+(1-k)^2=r^2\Rightarrow 2+k^2-2k=r^2.....(2)$

$Eq. (1) - Eq.(2)$ We get

$2k-1=0$

Solving these equations, we get,

$k=\dfrac{1}{2}$ and

$r^2=\dfrac{5}{4}$

Thus, the required equation of the circle is :

$x^2+(y-\dfrac{1}{2})^2=\dfrac{5}{4}$ i.e.,

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

State whether the following statements are true or false.
The equation

$x^{2}+y^{2} + 2x -10y + 30 = 0$ represents the equation of a circle.

0%
True
0%
False

Explanation

The radius of this circle =

$\sqrt{1^{2} + (-5)^{2} -30} = \sqrt{(-4)}$ which is imaginary

The equation of the ellipse whose equation of directrix is

$3x+4y-5=0$ , coordinates of the focus are

$(1,2)$ and the eccentricity is

$\dfrac{1}{2}$ is

$91x^2+84y^2-24xy-170x-360y+475=0$

0%
True
0%
False

Explanation

Let

$P(x,y)$ be any point on the ellipse and PM be the perpendicular from P upon the directrix

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

Then by the definition,

$\dfrac{SP}{PM}=e$

$SP=e.PM$

$\sqrt{(x-1)^2+(y-2)^2}=\dfrac{1}{2}|\dfrac{3x+4y-5}{\sqrt{3^2+4^2}}|$

$(x-1)^2+(y-2)^2=\dfrac{1}{4}. \dfrac{(3x+4y-5)^2}{25}$

$100(x^2+y^2-2x-4y+5)=9x^2+16y^2+24xy-30x-40y+25$

$91x^2+84y^2-24xy-170x-360y+475=0$ is the equation of the ellipse.

Centres of the three circles

${x}^{2}+{y}^{2}-4x-6y-14=0$

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

${x}^{2}+{y}^{2}-10x-16y+7=0$ . The centres of the circles are:

0%
are the vertices of a right angle
0%
the vertices of an isosceles triangle which is not regular
0%
vertices of a regular triangle
0%
are collinear

Explanation

${x^2} + {y^2} - 4x - 6y - 14 = 0$ ---(i)

${x^2} + {y^2} + 2x + 4y - 5 = 0$ ---(ii)

${x^2} + {y^2} - 10x - 16y + 7 = 0$ ---(iii)

centre of the circle (i)

${c_1} = \left( { - g, - f} \right) = \left( {2,3} \right)$

centre of the circle (ii)

${c_2} = \left( { - g, - f} \right) = \left( { - 1, - 2} \right)$

centre of the circle (iii)

${c_3} = \left( { - g, - f} \right) = \left( {5,8} \right)$

Now,

${c_1}{c_2} = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {3 + 2} \right)}^2}} = \sqrt {34}$

${c_2}{c_3} = \sqrt {{{\left( {5 + 1} \right)}^2} + {{\left( {8 + 2} \right)}^2}} = 2\sqrt {34}$

${c_1}{c_3} = \sqrt {{{\left( {5 - 2} \right)}^2} + {{\left( {8 - 3} \right)}^2}} = \sqrt {34}$

Since

${c_1}{c_2} + {c_1}{c_3} = {c_2}{c_3}$

Therefore

${c_1}$ ,

${c_2}$ and

${c_3}$ collinear.

The radius of the circle centred at

$(4,5)$ and passing through the centre of the circle

${x}^{2}+{y}^{2}+4x+6y-12=0$ is

0%
$2\sqrt {5}$
0%
$2\sqrt {10}$
0%
$3\sqrt {5}$
0%
$3\sqrt {10}$

Explanation

Let the center

$A(4,5)$

$x^2+y^2+4x+6y-12=0$

On comparing with it

$x^2+y^2+2gx+2fy+c=0$

$g=2,f=3,c=-12$

center of the circle

$=(-g,-f)=(-2,-3)$

Then required radius of the circle \

$=\sqrt{(4+2)^2+(5+3)^2}=10$

Centre of circle whose normal's are

$x^{2}-2xy-3x+6y=0$ , is

0%
$\left(3,\ \dfrac{3}{2}\right)$
0%
$\left(3,\ -\dfrac{3}{2}\right)$
0%
$\left(\dfrac{3}{2},\ 3\right)$
0%
$None\ of\ these$

Explanation

${x}^{2} - 2xy - 3x + 6y = 0$

$\Rightarrow \; \left( x - 3 \right) \left( x - 2y \right) = 0$

Hence,

$x = 3$ and

$x = 2y$ are two normals.

The intersection point of these two normals will be the centre of the circle.

$\therefore$ for

$x = 3$ ,

$\Rightarrow$

$y = \cfrac{x}{2} = \cfrac{3}{2}$

Hence, the intersection point is

$\left( 3, \cfrac{3}{2} \right)$ .

Hence, the centre of the given circle is

$\left( 3, \cfrac{3}{2} \right)$ .

The centre of the circle

$x^2+y^2+10x-20y+100=0$ is

0%
$(5,10)$
0%
$(-5,10)$
0%
$(-5,-10)$
0%
$(5,-10)$

Explanation

Given the equation of the circle is

$x^2+y^2+10x-20y+100=0$

or,

$x^2+10x+25+y^2-20y+100=25$

or,

$(x+5)^2+(y-10)^2=5^2$

From this equation it is clear that the centre is

$(-5,10)$ .

The length of the diameter of the circle

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

0%
$9$
0%
$3$
0%
$4$
0%
$6$

Explanation

Consider the given circle equation.

${{x}^{2}}+{{y}^{2}}-4x-6y+4=0$ …… (1)

We know that the general equation of circle,

${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ …… (2)

On comparing equation (1) and (2), we get

$2g=-4\Rightarrow g=-2$

$2f=-6\Rightarrow f=-3$

$c=4$

So, the centre of the circle

$C=\left( -g,-f \right)$

$C=\left( 2,3 \right)$

We know that the radius of circle,

$r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$

$r=\sqrt{{{2}^{2}}+{{3}^{2}}-4}$

$r=\sqrt{4+9-4}$

$r=\sqrt{9}=3$

So, the diameter of this circle

$d=2r=6$

Hence, this is the answer.

which of the following equations represents a parabola

0%
${\left( {x - y} \right)^3} = 3$
0%
$\frac{x}{y} - \frac{y}{x} = 0$
0%
$\frac{x}{y} + \frac{4}{x} = 0$
0%
${\left( {x + y} \right)^2} + 3 = 0$

Explanation

We know that the general equation of parabola is

${{y}^{2}}=4ax$

${{y}^{2}}=-4ax$

${{x}^{2}}=4ax$

${{x}^{2}}=-4ax$

From option (a),

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

It is not represent the parabola.

From option (b),

$\dfrac{x}{y}-\dfrac{y}{x}=0$

${{x}^{2}}={{y}^{2}}$

It is not represent the parabola.

From option (c),

$\dfrac{x}{y}+\dfrac{4}{x}=0$

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

${{x}^{2}}=-4y$

So, this is represented the parabola.

From option (d),

${{\left( x+y \right)}^{3}}+3=0$

It is not represent the parabola.

Hence, option c represents the parabola.

The locus of a point which is at a constant distance 5 from the fixed point

$(2,3)$ is:

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

Explanation

Let $\left( {x,y} \right)$ be point then ATQ

$\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2} = 5}$

${\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = 25$

${x^2} + {y^2} + 4 + 9 - 4x - 6y = 25$

$\Rightarrow {x^2} + {y^2} - 4x - 6y - 12 = 0$

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

Explanation

Let the equation be

$x^2 + y^2 + fx + gy + d = 0$ with centre =

$(-\cfrac{-f}{2} , -\cfrac{g}{2})$

Hence,

$f = -4 , g = 2$

Equation becomes =

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

Putting x = 3 y = 6 in the above equation we get

$9+36-12+12+d = 0$

$\implies d = -45$

$equation = x^2+y^2-4x+2y-45 = 0$

Find the equation of the circle :
Centered at

$(3,-2)$ with radius

$4$ .

0%
$x^2+y^2+6x-4y=3$
0%
$x^2+y^2-6x+4y=3$
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$x^2+y^2-3x+2y=-3$
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$x^2+y^2+3x-2y=-3$

Explanation

The center radius form equation of circle is

$(x-h)^2+(y-k)^2=r^2$ , where

$(h,k)$ is the center and

$r$ is the radius.

Here,

$(h,k)\equiv(3,-2)$ and

$r=4$

$\therefore$ Equation of circle: -

$(x-3)^2+(y-(-2))^2=4^2\Rightarrow x^2+9-6x+y^2+4+4y=16\Rightarrow x^2+y^2-6x+4y=3$

The equation to the circle with centre

$(2,1)$ and touches the line

$3x+4y-5$ is ?

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

Explanation

$\perp$ distance of pt.

$(2, 1)$ from line

$3x+4y-5$ is radius(r)

$\Rightarrow r=\dfrac{|6+4-5|}{5}=\dfrac{5}{5}=1$

$\Rightarrow$ Equation of circle is

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

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

$\Rightarrow$ Option C is correct.

If the vertices of a triangle are

$(2, -2), (-1, -1)$ and

$(5, 2)$ then the equation of its circumcircle is?

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$x^2+y^2+3x+3y+8=0$
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$x^2+y^2-3x-3y-8=0$
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$x^2+y^2-3x+3y+8=0$
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None of these

Explanation

To find circumcentre we write the equation of perpendicular bisectors of two sides and find their intersection,

3x-y-3=0 and 6x+8y-21=0

Their intersection point is ( $\dfrac{3}{2}$ , $\dfrac{3}{2}$ )

Radius of circumcircle = Distance of ( $\dfrac{3}{2}$ , $\dfrac{3}{2}$ )

from (2,-2) or any other vertex = $\dfrac{5}{\sqrt{2}}$

So equation of circle = ( $x-\frac{3}{2})^2$ +( $y-\frac{3}{2})^2$ = $\dfrac{25}{2}$ which corresponds to B option .

Length of the latus rectum of the hyperbola

$xy=c^{2}$ , is

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$2c$
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$\sqrt{2}c$
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$2\sqrt{2}c$
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$4c$

Explanation

$Given$

$equation\text{ of rectangular hyperbola is }$

$\text{xy=}{{\text{c}}^{2}}\,\,............\left( i \right)$

$We\text{ know that in a rectangular hyperbola the length of transverse axis, length of conjugate axis}$

$\text{and length of latus rectum all are equal and it is along}$

$\text{y=x}\,\,\,\,\,\,...........\left( ii \right)$

$so\text{ from }\left( i \right)$

$x\cdot x={{c}^{2}}$

$taking\,square\,root\,both\,sides$

$x=\pm c$

$Hence\,from\left( ii \right)$

$y=\pm c\,\,$

$\,Hence\,\left( x,y \right)=\left( c,c, \right)and\,\left( -c.-c \right)$

$We\text{ know}\,\text{distance between two points}\left( x_{1}^{{}},y_{1}^{{}} \right)and\left( x_{2}^{{}},y_{2}^{{}} \right)is$

$d=\sqrt{{{\left( x_{2}^{{}}-x_{1}^{{}} \right)}^{2}}+{{\left( y_{2}^{{}}-y_{1}^{{}} \right)}^{2}}}$

$Hence\,dis\tan ce\,between\,\left( c,c, \right)and\,\left( -c.-c \right)$

$d=\sqrt{{{\left( -c-c \right)}^{2}}+{{\left( -c-c \right)}^{2}}}$

$\,\,\,\,=\sqrt{4{{c}^{2}}+4{{c}^{2}}}$

$\,\,\,=2\sqrt{2}\,c$

For what value of

$k$ , does the equation

$9{x^2} + {y^2} = k\left( {{x^2} - {y^2} - 2x} \right)$ represents equation of a circle?

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$1$
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$2$
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$-1$
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$4$

Explanation

$9{ x }^{ 2 }-k{ x }^{ 2 }+{ y }^{ 2 }+k{ y }^{ 2 }+2kx=0\\ { x }^{ 2 }(9-k)+{ y }^{ 2 }(1+k)+2kx=0$

for circle

$9-k=1+k$

So,

$k=4$

The length of latus recturn of the hyperbola

${ 9x }^{ 2 }-16{ y }^{ 2\quad }+72x-32y-16=0$

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$\dfrac { 9 }{ 2 }$
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$\dfrac {- 9 }{ 2 }$
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$\dfrac { 32 }{ 3 }$
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$\dfrac { -32 }{ 2 }$

Equation of the hyperbola with eccentricity

$\cfrac{3}{2}$ and foci at

$(\pm 2, 0)$ is

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$\cfrac{x^2}{4}-\cfrac{y^2}{5}=\cfrac{4}{9}$
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$\cfrac{x^2}{9}-\cfrac{y^2}{4}=\cfrac{4}{9}$
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$\cfrac{x^2}{4}-\cfrac{y^2}{9}=1$
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none of these

The centre of the circle given by

$\mathbf { r } \cdot ( \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } ) = 15 \text { and } | \mathbf { r } - ( \mathbf { j } + 2 \mathbf { k } ) | = 4 ,$

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$( 0,1,2 )$
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$( 1,3,4 )$
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$( - 1,3,4 )$
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None of these

Explanation

$\begin{array}{l} The\, equation\, of\, line\, pas\sin g\, through\, centre \\ \vec { i } +2\vec { k } \, and\, normal\, to\, the\, given\, plane \\ \vec { r } =\vec { j } +2\vec { k } +\lambda \left( { \vec { i } +2\vec { j } +2\vec { k } } \right) \, \, \, \, \, \, -----\left( 1 \right) \\ This\, plane\, meets\, the\, plane\, at\, a\, po{ { int } }\, for\, which \\ \left[ { \left( { \vec { j } +2\vec { k } } \right) +\lambda \left( { \vec { i } +2\vec { j } +2\vec { k } } \right) } \right] \left( { \vec { i } +2\vec { j } +2\vec { k } } \right) =15 \\ \Rightarrow 6+9\lambda =15 \\ \Rightarrow 9\lambda =9 \\ \therefore \lambda =1 \\ substituting\, value\, of\, \lambda \, in\, eq{ u^{ n } }\left( 1 \right) \\ \vec { r } =\vec { j } +2\vec { k } +\vec { i } +2\vec { j } +2\vec { k } \\ \Rightarrow \vec { r } =\vec { i } +3\vec { j } +4\vec { k } \\ \therefore Centre\, is\, \left( { 1,\, 3,\, 4 } \right) \end{array}$

Hence, option

$B$ is the correct answer.

The equation of the circle passing through (2,0) and (0,4) and having the minimum radius is

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${ x }^{ 2 }+{ y }^{ 2 }=20$
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${ x }^{ 2 }+{ y }^{ 2 }-2x-4y=0$
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${ (x }^{ 2 }+{ y }^{ 2 }-4)+\lambda ({ x }^{ 2 }+{ y }^{ 2 }-16)=0$
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N.O.T.

Explanation

Let

$(x-h)^2+(y-k)^2=r^2$ …………..

$(1)$

where (h, k) is the centre of the circle and r is the radius.

Let us draw a line

$A(2, 0)$ and

$B(4, 0)$ . If the centre lies outside the line AB then it is a chord.

The diameter of the circle is bigger than the chord. Suppose AB is the diameter of the circle then the centre of the circle must be a midpoint of AB.

$\therefore$ Using section formula

$h=\dfrac{2+0}{2}=\dfrac{2}{2}=1$

$k=\dfrac{4+0}{2}=\dfrac{4}{2}=2$

$\therefore$ Coordinates of the centre

$=(h, k)$

$=(1, 2)$

Equation

$(1)$ becomes,

$(x-1)^2+(y-2)^2=r^2$ …………….

$(2)$

Since

$(2)$ passes through

$(2, 0)$ , equation

$(2)$ can be written as,

$(2-1)^2+(0-2)^2=r^2$

$1^2+2^2=r^2$

$1+4=r^2$

$r^2=5$

$r=\sqrt{5}$

Equation of the circle with minimum radius is

$\therefore (x-1)^2+(y-2)^2=5$

$\Rightarrow x^2+1-2x+y^2+4-4y=5$

$\Rightarrow x^2+y^2-2x-4y+5-5=0$

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

1209354_1300403_ans_ce254d19d4374c62a6072d6bf1799c07.JPG

Which is not represented by quadratic equation ?

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Circle
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Straight line
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Parabola
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Hyperbola

Explanation

Equation of circle is given by:

$x^2+y^2+2gx+2fy+c$

Equation of straight line is given by:

$y=mx+c$

Equation of parabola is given by:

$(y-k)^2=4p(x-h)^2$

Equation of hyperbola is given by:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

Hence, we can see from the above equations that only equation of a straight line is not represented by quadratic equation.

1176114_1327206_ans_3e4b5ba0f4f84fbd92e82deff706951f.jpg

Find the area of

$x^2+y^2=49$

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154
0%
49
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88
0%
None

Explanation

Th eequation

$x^2+y^2=49$ describes a circle with

$7$ as radius

So the area is given as

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

The equation

${ x }^{ 2 }+{ y }^{ 2 }=9$ meets x-axis at

0%
$(\pm 3,0)$
0%
$(\pm 9,0)$
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$(\pm 1,0)$
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(-5/14, 5/14)

Explanation

$x^2+y^2=9$

The equation meets x-axis if the y-coordinate is

$0$

$x^2+0=9\\x^2=9\\x=\sqrt9\\x=\pm 3$

So the point is

$(\pm 3,0)$

If the the hyperbola

$\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ passses though

$(4,3)$

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${ b }^{ 2 }=3$
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${ b }^{ 2 }=9$
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${ b }^{ 2 }=4$
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${ b }^{ 2 }=100$

Explanation

Equation of hyer bola is

$\dfrac{x^2}{4}-\dfrac{y^2}{b^2}=1$ passes through

$(4,3)$

$\dfrac{4^2}{4}-\dfrac{3^2}{b^2}=1\\4-\dfrac{9}{b^2}=1\\ \dfrac{9}{b^2} =3\implies b^2=3$

The equation

$\dfrac {x^{2}}{2-r}+\dfrac {y^{2}}{r-5}+1=0$ represents an ellipse, if

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$r > 2$
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$2 < r < 5$
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$r > 5$
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$r \in (2,5)$

Explanation

Given

$\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$ represents a ellipse

$\implies \dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}=-1$

$\implies \dfrac{x^2}{r-2}+\dfrac{y^2}{5-r}=1$

Since this equation is an ellipse so

$r-2>0,5-r>0\implies 2<r<5$

Find the value of a if

$y^2=4ax$ pases through

$(8,8)$

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2
0%
4
0%
8
0%
None

Explanation

Given point

$(8,8)$

Given equation

$y^2=4ax$

$\implies 8^2=4a(8)\\64=32a\\a=\dfrac{64}{32}\\a=2$

Find the Center of circle

$x^2+y^2-4x-8x+25=0$

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$(2,4)$
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$(-2,-4)$
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$(4,2)$
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$(-4,-2)$

Explanation

The general equation of center of circle

$x^2+y^2+2gx+2fy+c=0$ is

$(-g,-f)$

So the center of circle

$x^2+y^2-4x,-8x+25=0$ is

$(2,4)$

The equation of the circle passing through

$(2,0)$ and

$(0,4)$ and having the minimum radius is ______________.

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

Explanation

Equation of circle with centre (a,b) and radius r,

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

For the circle to have minimum radius, given points must be diametric points,

So, Centre of circle =

$(\dfrac{2+0}{2},\dfrac{0+4}{2}) = (1,2)$

Radius of circle =

$\sqrt{(1-0)^2 + (2-4)^2} = \sqrt{5}$

So, The equation of circle is

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

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

Radius of the circle

$2x^2+2y^2+8x+4y-3=0$ is

0%
$\sqrt{17}$
0%
$\sqrt{23}$
0%
$\sqrt{\dfrac{13}{2}}$
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$\sqrt{\dfrac{11}{2}}$

Explanation

Formula,

$r=\sqrt{g^2+f^2-c}$

From given, we have,

$2g=8\Rightarrow g=4$

$2f=4\Rightarrow f=2$

$c=-3$

$r=\sqrt{4^2+2^2-(-3)}=\sqrt{23}$

A rod

${A}{B}$ of length

$4$ units moves horizontally with its left end

$\mathrm{A}$ always on the circle

$x^{2}+y^{2}-4x-18y-29=0$ then the locus of the other end

$\mathrm{B}$ is

0%
$x^{2}+y^{2}-12x-8y+3=0$
0%
$x^{2}+y^{2}-12x-18y+3=0$
0%
$x^{2}+y^{2}+4x-18y-29=0$
0%
$x^{2}+y^{2}-4x-16y+19=0$

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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