MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 8

The intercept on the line

$y = x$ by the circle

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

$AB$ . The 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%
${x^2} + {y^2} - 3x + y = 0$
0%
None of the above

The equation of the circle passing through the points of intersection of the lines

$2x+y=0,\ x+y+3=0$ and

$x-2y=0$ is

0%
$x^{2}+y^{2}-x+7y=0$
0%
$x^{2}+y^{2}+x-7y=0$
0%
$x^{2}+y^{2}-x-7y=0$
0%
$x^{2}+y^{2}+x+7y=0$

Length of the latus rectum of the parabola

$25 [(x- 2)^{2}+(y-3)^{2}]=(3x-4y+7)^2$ is :

0%
4
0%
2
0%
1/5
0%
2/5

Explanation

$The\, given\, equation\, is\, \\ 25\left[ { { { \left( { x-2 } \right) }^{ 2 } }+{ { \left( { y-3 } \right) }^{ 2 } } } \right] ={ \left( { 3x-4y+7 } \right) ^{ 2 } } \\ Here, \\ focus\, of\, parabola\, \left( { 2,3 } \right) \\ Directrix \\ 3x-4y+7$

$\\ Now,\, we\, know\, distan ce\, between\, directrix\, and\, \, focus\, is\, \, half\, of\, latus\, rectum. \\ so,\, distan ce\, between\, them\Rightarrow L=\left| { \dfrac { { 3\times 2-4\times 3+7 } }{ { \sqrt { { 3^{ 2 } }+{ 4^{ 2 } } } } } } \right| \\ L=\left| { \dfrac { { 6-12+7 } }{ { \sqrt { 25 } } } } \right| \\ \therefore L=\left| { \dfrac { 1 }{ 5 } } \right| \\ Then\, latus\, rectum=2L \\ =2\times \dfrac { 1 }{ 5 } \\ =\dfrac { 2 }{ 5 }$

$\\ Hence,\, the\, option\, D\, is\, the\, correct\, answer.$

The equation of the circle which touches both the axes and the line

$\frac { x }{ 3 } +\frac { y }{ 4 } =1$ and lies in the first quadrant is

$(x-c)^{ 2 }+(y-c)^{ 2 }={ c }^{ 2 }$ where c is

0%
1
0%
2
0%
4
0%
6

The length of the latus rectum of the parabola

$x^{2}-4x-8y+12=0$ is-

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

Explanation

We have

${x^2} - 4x - 8y + 12 = 0$

$\Rightarrow {\left( {x - 2} \right)^2} + 8 - 8y = 0\,\,\,\,\,\left( {by\,taking\,common\,x} \right)$

$\Rightarrow {\left( {x - 2} \right)^2} = 8\left( {y - 1} \right)$

Comparing with

$X^2=L.R\times Y$

So, Length of latus rectum

$=8$

Hence, the option

$(C)$ is the correct answer.

The equation of the circle of a radius

$5$ in the first quadrant which touches the x-axis and the line

$3x-4y=0$ is

0%
$x^{2}+y^{2}-24x-y-25=0$
0%
$x^{2}+y^{2}-20x-12y+144=0$
0%
$x^{2}+y^{2}-16x-18y+64=0$
0%
$x^{2}+y^{2}-30x-10y+255=0$

Equation of circle touching

$x=0, y=0$ and

$x=4$ is

0%
$4(x^{2}+y^{2})-16x-16y+16=0$
0%
$4(x^{2}+y^{2})-12x-12y+12=0$
0%
$4(x^{2}+y^{2})-8x-8y+4=0$
0%
$x^{2}+y^{2}-x-y-1=0$

Explanation

$\textbf{Step -1: Finding the centre and radius from the given values.}$

$\text{We are given that}$

$(h,k)=(2,2),r=2$

$\textbf{Step -2: Finding the equation of the circle}$

$\text{Equation of a circle whose centre is}$

$(h,k)$

$\text{and radius is}$

$r$

$\text{is}$

$\Rightarrow (x-h)^2+(y-k)^2=r^2$

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

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

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

$\text{Multiplying throughout by}$

$4,$

$\Rightarrow 4(x^2+y^2)-16x-16y+16=0$

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

The area of the triangle formed by the tangent and the normal to the parabola

${ y }^{ 2 }=4ax,$ both drawn at the same end of the latus rectum and the axis of the parabola is

0%
$2\sqrt { 2 } { a }^{ 2 }$
0%
$2{ a }^{ 2 }$
0%
$4{ a }^{ 2 }$
0%
None of these

Find the equation of a circle whose center is (2,-1) and radius is 3

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

Explanation

Equation of circle with center

$(h,k)$ and radius

$r$ is written as

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

then,

Equation of circle with center

$(2,-1)$ and radius

$3$ is written as

$\Rightarrow (x-2)^2+(y+1)^2=3^2$

$\Rightarrow x^2+4-4x+y^2+1+2y=9$

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

The vertex and focus of a parabola are

$(-2,2),(-6,6)$ . Then its length of latus rectum is

0%
$8\sqrt {2}$
0%
$4\sqrt {2}$
0%
$16\sqrt {2}$
0%
$12\sqrt {2}$

Explanation

Distance between focus and vertex

$\begin{array}{l} focus\left( { -6,6 } \right) ,\, vertex\left( { -2,2 } \right) \\ a=\sqrt { { { \left( { -6+2 } \right) }^{ 2 } }{ { \left( { 6-2 } \right) }^{ 2 } } } \\ a=\sqrt { 16+16 } =4\sqrt { 2 } \end{array}$

length latus vectum

$\begin{array}{l} =4a=4\times 2\sqrt { 2 } \\ =16\sqrt { 2 } \end{array}$

The equation of the tangent to the curve y = 2sinx + sin2x at

$x=\frac { \pi }{ 3 }$ on it is

0%
$(2,3)$
0%
$y+\sqrt { 3 } =0$
0%
$2t-3=0$
0%
$2y-3\sqrt { 3 } =0$

Explanation

${ (x-2) }^{ 2 }+{ (y-3) }^{ 2 }=0\\ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+9-6y=0\\ { x }^{ 2 }+{ y }^{ 2 }-4x+6y+13=0$

radius

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

$g=-2 \quad f=-3 \quad c=13$

$r=\sqrt { { (-2) }^{ 2 }+{ (-3) }^{ 2 }-13 } =\sqrt { 13-13 } =0$

radius of the circle is zero .

Hence diameter

$=0$

i.e., it is a point circle with centre at

$(2,3)$ .

The order of the differential equation of the family of parabolas whose length of latus rectum is fixed and axis is the X-axis

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

Explanation

$\begin{array}{l} Equation\, can\, be\, { \left( { y-k } \right) ^{ 2 } }=a\left( { x-h } \right) \, \, \, a=fixed \\ differentiating\, this\, we\, get \\ 2\left( { y-k } \right) .y'=ax \\ \left( { y-k } \right) .{ y^{ n } }+y{ '^{ 2 } }=\frac { a }{ 2 } \\ \frac { { ax } }{ { 2y' } } .y''+y{ '^{ 2 } }=\frac { a }{ 2 } \\ ax.{ y^{ n } }+2y{ '^{ 3 } }=ay' \\ \therefore order\, =2 \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

${ \cos }^{ 4 }\dfrac { \pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 3\pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 5\pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 7\pi }{ 8 } =$

0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { 3 }{ 2 }$
0%
$\cfrac { 1 }{ 4 }$
0%
$\cfrac { 3 }{ 4 }$

Explanation

$\textbf{Step 1: Simplify the given expression.}$

$\text{Consider, } \cos^4\dfrac{\pi}{8} + \cos^4\dfrac{3\pi}{8} + \cos^4\dfrac{5\pi}{8} + \cos^4\dfrac{7\pi}{8}$

$= \cos^4\dfrac{\pi}{8} + \cos^4\left(\dfrac{\pi}{2}-\dfrac{\pi}{8}\right) + \cos^4 \left ( \dfrac{\pi}{2} + \dfrac{\pi}{8} \right ) + \cos^4 \left ( \pi - \dfrac{\pi}{8} \right )$

$= \cos^4\dfrac{\pi}{8} + \sin^4\dfrac{\pi}{8} + \left (-\sin\dfrac{\pi}{8} \right )^4 + \left ( -cos\dfrac{\pi}{8} \right )^4$

$= 2\left [ \cos^4\dfrac{\pi}{8} + \sin^4\dfrac{\pi}{8} \right]$

$\textbf{Step 2: Use multiple angle formula to get final result. }$

$= 2 \left \{ \left ( \cos^2\dfrac{\pi}{8} + \sin^2\dfrac{\pi}{8} \right )^2 - 2\cos^2\dfrac{\pi}{8} \cdot \sin^2\dfrac{\pi}{8} \right \}$

$= 2 \left ( 1^2 - \dfrac{1}{2} \left ( 2\sin\dfrac{\pi}{8}\cos\dfrac{\pi}{8} \right )^2 \right )$

$= 2\left [ 1^2 - \dfrac{1}{2} \times \left ( \sin\dfrac{\pi}{4} \right )^2 \right ]$

$[\because \boldsymbol{\sin 2A=2\sin A\cos A}]$

$= 2\left ( 1^2 - \dfrac{1}{2} \times \left ( \dfrac{1}{\sqrt 2} \right )^2 \right )$

$= 2\left [1^2 - \dfrac{1}{4} \right ]=\dfrac{3}{2}$

$\textbf{Hence, Option 'B' is correct.}$

Length of the latusrectum of the hyperbola

$xy-3x-4y+8=0$ is

0%
$4$
0%
$4\sqrt{2}$
0%
$8$
0%
$None\,of\,these$

Explanation

Hyperbola

$=xy-3x-4y+8=0$

(add & sub

$12$ )

$x(y-3)-4(y-3)+8-12=0$

$(x-4)(y-3)=4$

It is in form XY

$=c^2$

where X

$=x-4$

$=y-3$

Latus rectum

$=\dfrac{2b^2}{a}=2(a)$ [rectangular hyperbola

$b=a$ ]

We know that

$a=\sqrt{2}.C$

L.R

$=2\sqrt{2}\cdot C$

$=2.2.\sqrt{2}$

$=4\sqrt{2}$ .

1214594_1317557_ans_91f2728245ea4e18bcee68fa61065281.jpeg

The latus rectum of the hyperbola

$16{x^2} - 9{y^2} = 144$ is-

0%
$\dfrac{13}{6}$
0%
$\dfrac{32}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{4}{3}$

Explanation

We have,

$16{x^2} - 9{y^2} = 144$

$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Here,

$a=3, b=4$

We know that the latus rectum

$=\dfrac{2b^2}{a}$

Therefore,

$=\dfrac{2\times 16}{3}$

$=\dfrac{32}{3}$

Hence, this is the answer.

The equation

$14x^{2}-4xy+11y^{2}-44x-58y+71=0$ represents

0%
a circle
0%
an ellipse
0%
a hyperbola
0%
a rectangular hyperbola

Length of the latus rectum of the parabola

$25\left[ {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2}} \right] = {\left( {3x - 4y + 7} \right)^2}$ is:

0%
$4$
0%
$2$
0%
$\dfrac {1}{5}$
0%
$\dfrac {2}{5}$

Length of the latus rectum of the hyperbola

$xy-3x-4y+8=0$

0%
$4$
0%
$4\sqrt{2}$
0%
$8$
0%
none of these

Explanation

$xy-3x-4y+8=0$

$x(y-3)-4(y-3)-12+8=0$

$(x-4)(xy-3)=4$

$xy=c^2$ form

Latus nectum

$=\dfrac{2b^2}{a}=2a=2\sqrt{2}c$

$=4\sqrt{2}$

Option B

$=4\sqrt{2}$ .

1243454_1327986_ans_fe9e39a1ec804122b6c5bc9025d79fd2.jpg

The line

$y=mx+c$ cut the circle

${x}^{2}+{y}^{2}={a}^{2}$ in the distinct point

$A$ and

$B$ . Equation of the circle having minimum radius that an be drawn through the points

$A$ and

$B$ is

0%
$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) +2c\left( y-mx-c \right) =0$
0%
$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) +c\left( y-mx-c \right) =0$
0%
$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) -2c\left( y-mx-c \right) =0$
0%
$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ 2a }^{ 2 } \right) -2c\left( y-mx-c \right) =0$

Three sides of a triangle have the equations

$L_{r} = y - m_r x - C_{r} = 0; r = 1, 2, 3$ . Then

$\lambda L_{2}L_{3} + \mu L_{3}L_{1} + \gamma L_{1}L_{2} = 0$ . where

$\lambda \neq 0, \mu \neq 0, \gamma \neq 0$ , is the equation of circumcircle of triangle if

0%
$\lambda (m_{2} + m_{3}) + \mu (m_{3} + m_{1}) + \gamma (m_{1} + m_{2}) = 0$
0%
$\lambda (m_{2}m_{3} - 1) + \mu (m_{3}m_{1} - 1) + \gamma (m_{1}m_{2} - 1) = 0$
0%
Both $(a)$ and $(b)$ hold together
0%
None of these

The equation of the circle passing through the points

$(4, 1), (6, 5)$ and having the centre on the line

$4x+y-16=0$ is

0%
${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+15=0$
0%
$15({ x }^{ 2 }+{ y }^{ 2 })-94x+18y+55=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-4x-3y=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+6x-4y=0$

Explanation

Given points,

$(4,1),(6,5)$

equation of circle

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

$\Rightarrow (4-h)^2+(1-k)^2=r^2$ ....(1)

$\Rightarrow (6-h)^2+(5-k)^2=r^2$ ....(2)

solving the above 2 equations, we get,

$h+2k=11$ ....(3)

given,

$4h+k=16$ .....(4)

solving the above 2 equations, we get,

$h=3,k=4$

substituting the above values in (1), we get,

$(4-3)^2+(1-4)^2=r^2$

$\therefore r=\sqrt{10}$

Hence, the equation is,

$(x-3)^2+(y-4)^2=(\sqrt {10})^2$

$x^2+y^2-6x-8y+15=0$

The length of latus rectum of the parabola

$4y^{2}+3x+3y+1=0$ is

0%
$\dfrac {4}{3}$
0%
$7$
0%
$12$
0%
$\dfrac {3}{4}$

Equation of the circle which is such that the lengths of the tangents to it from the points

$( 1,0 ) , ( 0,2 )$ and

$( 3,2 )$ are

$1 , \sqrt { 7 }$ and

$\sqrt { 2 }$ respectively is

0%
$6 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
0%
$9 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
0%
$3 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
0%
$x ^ { 2 } + y ^ { 2 } - x + y + 1 = 0$

Which of the following is the equation of a circle?

0%
$x^2 + 2y^2 - x + 6 = 0$
0%
$x^2 - y^2 + x + y + 1 = 0$
0%
$x^2 - y^2 + xy + 1 = 0$
0%
$3(x^2 + y^2) + 5x + 1 = 0$

The latus rectum of an ellipse is a line

0%
Through a focus
0%
Through the centre
0%
Perpendicular to major axis
0%
Parallel to minor axis

The equation to the circle which touches the axis of y at the origin and passes through

$(3, 4)$ is?

0%
$2(x^2+y^2)-3x=0$
0%
$3(x^2+y^2)-25x=0$
0%
$4(x^2+y^2)-25y=0$
0%
$4(x^2+y^2)-25x+10=0$

The image of the circle

$(x - 3)^2 + (y - 2)^2 = 1$ in the line mirror

$ax + by = 19$ is

$(x - 1)^2 + (y - 16)^2 = 1$ then values of (a, b) is

0%
(1,1)
0%
(1,1)
0%
(1,1)
0%
(-1,-1)

The radius of circle

$x^2+y^2-6x-8y=0$

0%
5
0%
4
0%
3
0%
2

Explanation

The radius of circel $x^2+y^2-6x-8y=0$ is $\sqrt{g^2+f^2-c}$

Here $g=-3,f=-4,c=0\implies r=\sqrt{(-3)^2+(-4)^2}\\\sqrt{9+16}=\sqrt {25}=5$

The equation of circle center at

$(0,0)$ and Radius

$8cm$

0%
$x^2+y^2=64cm$
0%
$x^2+y^2=8$
0%
$x^2+y^2=16$
0%
$x^2+y^2=4$

Explanation

The equation of circle is

$x^2+y^2=r^2\\x^2+y^2=8^2\\x^2+y^2=64$

A circle of radius

$'5'$ touches the coordinate axes in the first quadrant. If the circle makes one complete roll on x-axis along the positive direction, then its equation in new position is

0%
${x}^{2}+{y}^{2}-10(2\pi+1)x-10y+100{\pi}^{2}+100\pi+25=0$
0%
${x}^{2}+{y}^{2}+10(2\pi+1)x-10y+100{\pi}^{2}+100\pi+25=0$
0%
${x}^{2}+{y}^{2}-10(2\pi+1)x+10y+100{\pi}^{2}+100\pi+25=0$
0%
${x}^{2}+{y}^{2}+10(2\pi+1)x+10y+100{\pi}^{2}+100\pi+25=0$

If the circle

$x^{2}+y^{2}=9$ passesthrough

$(2,c)$ then

$c$ is equal to

0%
$\sqrt5$
0%
$\sqrt 6$
0%
$\sqrt 3$
0%
$\sqrt 7$

Explanation

The equation of circle

$x^2+y^2=9$

The point is

$(2,c)$

$\implies 2^2+c^2=9\\4+c^2=9\\c^2=9-4\\c^2=5\\c=\sqrt 5$

The equation of the circle of radius

$5$ with centre on x-axis and passing through the point

$(2,3)$ is

0%
$x^{2}+y^{2}-12x+11=0$
0%
$x^{2}+y^{2}-4x-21=0$
0%
$x^{2}+y^{2}+12x+11=0$
0%
$x^{2}+y^{2}-4x+21=0$

Explanation

For a circle with center (a,b) and radius r,

Equation is

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

Let the center of given circle be (k,0).

Then, equation of circle,

$\rightarrow$

$(x-k)^2 + (y-0)^2 = 5^2$

Since circle passes through (2,3),

$\rightarrow$

$(2-k)^2 + 3^2 = 5^2$

Solving we get, k = {6,-2}

So, Possible equations are,

(i)

$(x-6)^2 + y^2 = 25$

$\rightarrow$

$x^2 + y^2 -12x +11 = 0$

(ii)

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

$\rightarrow$

$x^2 + y^2 + 4x -21 = 0$

Hence, (A) and (B) are correct answers.

The equation of the circle which touches the axis of y at a distance

$+4$ from the origin and cuts off an intercept

$6$ from the

$+ve$ direction of x-axis is

0%
$x^{2}+y^{2}-10x\pm8y-16=0$
0%
$x^{2}+y^{2}+10x\pm8y+16=0$
0%
$x^{2}+y^{2}-10x\pm8y+16=0$
0%
$x^{2}+y^{2}-8x\pm10y-16=0$

Explanation

For a circle with center (a,b) and radius r,

Equation is

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

There can be two scenarios for the given conditions as depicted in above figures.

For (A),

$\triangle OMN$ , using pythagoras theorem,

$ON = \sqrt{OM^2 + MN^2} = \sqrt{4^2 + 3^2} = 5$

So, radius of circle = 5.

And, center = (5,4)

Thus, Equation of cricle

$\rightarrow$

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

$\rightarrow$

$x^2 + y^2 -10x -8y +16 = 0$

Similarly, For (B).

Equation of circle,

$\rightarrow$

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

$\rightarrow$

$x^2 + y^2 -10x +8y +16 = 0$

1824029_1416220_ans_4292957a0e9c4851b91971a01511f8df.jpg

The equation of the circle passing through the origin and making intercept

$4,5$ on the positive coordinates axes is

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

Explanation

For a circle with center (a,b) and radius r,

Equation is

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

Their can be two scenarios for given conditions as depicted in above figures.

For (A),

Center =

$(2,\dfrac{5}{2})$

Radius =

$\sqrt{2^2+(\dfrac{5}{2}})^2 = \dfrac{\sqrt{41}}{2}$

So, Equation of circle,

$\rightarrow$

$(x-2)^2 + (y-\dfrac{5}{2})^2 = \dfrac{41}{4}$

$\rightarrow$

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

Similarly, For (B),

Equation of circle,

$\rightarrow$

$(x-\dfrac{5}{2})^2 + (y-2)^2 = \dfrac{41}{4}$

$\rightarrow$

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

Hence, (B) is the correct answer.

1824280_1416460_ans_2da2aad0fb454884a01bc4367e93216c.jpg

The equation of the circle, the end points of whose diameter are the centre of the circles

$x^{2}+y^{2}+6x-14y=1$ and

$x^{2}+y^{2}-4x+10y=2$ is

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

Explanation

The general equation of circle is

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

where,

Center =

$(-g, -f)$

Radius =

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

So,

For

$x^2 + y^2 + 6x - 14y =1$ ,

Center =

$(-3,7)$

And For

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

Center =

$(2,-5)$

Since centers of given circles are diametric ends of new circle,

For new circle,

Center =

$(\dfrac{-3+2}{2}, \dfrac{7-5}{2}) = (\dfrac{-1}{2}, 1)$

Radius =

$\dfrac{\sqrt{(-3-2)^2 + (7+5)^2}}{2} = \dfrac{13}{2}$

Hence, Equation of new circle

$\rightarrow (x+\dfrac{1}{2})^2 + (y-1)^2 = \dfrac{169}{4}$

$\rightarrow x^2 + y^2 + x -2y - 41 = 0$

The equation of a circle with centre at

$(2,-3)$ and the circumference is

$10 \pi$ units is

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

Explanation

For a circle with center (a,b) and radius r,

Equation is

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

First let's find radius of circle,

$\rightarrow$ Circumference =

$2\times \pi \times r$

$\rightarrow$

$10 \times \pi$ =

$2 \times \pi \times r$

$\rightarrow$

$r = 5$

Equation of circle,

$\rightarrow$

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

$\rightarrow$

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

Equation having circle centre

$(5, 2)$ and which passes through the point

$(1, -1)$ is

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$x^2+y^2-10x-4y-4=0$
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$x^2+y^2+10x+4y+4=0$
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$x^2+y^2-10x-4y-2=0$
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$x^2+y^2-10x-4y+4=0$

Explanation

equation of circle,

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

$C=(5,2),P=(1,-1)$

$\Rightarrow (1-5)^2+(-1-2)^2=r^2$

$16+9=r^2$

$\therefore r=5$

Now,

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

$x^2+25-10x+y^2-4-4y=25$

$x^2+y^2-10x-4y-4=0$

Equation of circle touching the line

$x + y = 4$ at

$( 1, 3)$ and intersecting the circle

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

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${ x }^{ 2 }+{ y }^{ 2 }-x+2y-15=0$
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${ x }^{ 2 }+{ y }^{ 2 }-x-y-6=0$
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${ 2x }^{ 2 }+2{ y }^{ 2 }-x+y-22=0$
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${ 2x }^{ 2 }+2{ y }^{ 2 }-x-9y+8=0$

Explanation

The general equation of circle touching the line lx + my + n = 0 at given point

$(x_1,y_1)$ is

$(x-x_1)^2+(y-y_1)^2 + \lambda(lx+my+n)=0$ where

$\lambda \in R$

For two circles with equation,

$x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and

$x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ ,

Condition of orthogonality

$\rightarrow 2g_1g_2 + 2f_1f_2 = c_1 + c_2$

So, for given condition equation will be,

$\rightarrow (x-1)^2 + (y-3)^2 + \lambda (x+y-4) = 0$

$\rightarrow x^2 + y^2 +(\lambda -2)x + (\lambda-6)y +( 10-4\lambda) = 0$

Using condition of orthoganility,

$2(2-\lambda)(0) + 2(6-\lambda)(0) = -4 + 10 - 4\lambda$

$\rightarrow \lambda = \dfrac{3}{2}$

So, Equation of circle,

$\rightarrow x^2 + y^2 - \dfrac{1}{2}x -\dfrac{9}{2}y + 4 = 0$

$\rightarrow 2x^2 + 2y^2 -x-9y + 8 =0$

Hence, (D) is the correct answer.

The parametric equation of the circle

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

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$x=\dfrac{-1}{2}+\cos \theta,y=\dfrac {-\sqrt {3}}{2}+\sin\theta$
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$x=-\dfrac {1}{2}+\cos \theta,y=\dfrac {\sqrt {3}}{2}+\sin\theta$
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$x=\dfrac {1}{2}+\cos \theta,y=\dfrac {-\sqrt {3}}{2}+\sin\theta$
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$x=\dfrac {1}{2}+\cos \theta,y=\dfrac {\sqrt {3}}{2}+\sin\theta$

Explanation

The parametric equation of circle with radius r and center (a,b) is,

x = a + rcos

$\theta$ , y = b + rsin

$\theta$

The general equation on the other side is,

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

Let's find the radius using equation

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

We can rewrite equation as,

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

Comparing this with general equation, we get,

$a = \dfrac{-1}{2}, b= \dfrac{-\sqrt{3}}{2}, r = 1$

Hence, the parametric equation of circle is,

x =

$\dfrac{-1}{2} + cos\theta$ , y =

$\dfrac{-\sqrt{3}}{2} + sin\theta$

The equation

$\sqrt{(x-3)^{2}+(y-1)^{2}}+\sqrt{(x-3)^{2}+(y-1)^{2}}=6$ represents :

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an ellipse
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a pair of straight lines
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a circle
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the line segment joining the point $(-3,1)$ to the point $(3,1)$

If the line

$x-1=0$ is the directrix of the parabola

${ y }^{ 2 }-kx+8=0$ , then one of the values of

$k$ is

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${ 1 }/{ 8 }$
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$8$
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$4$
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${ 1 }/{ 4 }$

Explanation

The parabola is

$y^2-kx+8=0$

$\Rightarrow$

$y^2=kx-8$

$=k\left(x-\dfrac{8}{k}\right)$

$=4AX$

Where

$4A=k$

Direction of the parabola is

$X+A=0$

Comparing it with

$x-1=0$ we get,

$\Rightarrow$

$1=\dfrac{8}{k}-\dfrac{k}{4}$

$\Rightarrow$

$4k=32-k^2$

$\Rightarrow$

$k^2+4k-32=0$

$\Rightarrow$

$k^2+8k-4k-32=0$

$\Rightarrow$

$k(k+8)-4(k+8)=0$

$\Rightarrow$

$(k+8)(k-4)=0$

$\therefore$

$k=-8,4$

$\therefore$ From the given options one of the value of

$k$ is

$4$

The equation of the circle with centre at

$(4, 3)$ and touching the line

$5x-12y-10=0$ is?

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$x^2+y^2-4x-6y+4=0$
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$x^2+y^2+6x-8y+16=0$
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$x^2+y^2-8x-6y+21=0$
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$x^2+y^2-24x-10y+144=0$

The locus of center of a variable circle touching the circle of radius

${ r }_{ 1 }and{ r }_{ 2 }$ extemally which also touch each other externally , is a conic of the eccentricity

$e$ .If

$\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =3+2\sqrt { 2 }$ then

${ e }^{ 2 }$ is

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2
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3
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4
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5

The equation of the latus rectum of the parabola

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

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3y = 2
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2y + 3 = 0
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2y = 3
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3y + 2 = 0

If two vertices of an equilateral triangle are

$A (-a, 0)$ and

$B(a, 0), a > 0$ and the third vertex C lies above x-axis then the equation of the circumcircle of

$\Delta ABC$ is

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$3x^2+3y^2-2\sqrt{3} ay =3a^2$
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$3x^2+3y^2-2 ay =3a^2$
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$x^2+y^2-2ay =a^2$
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$x^2+y^2-\sqrt{3} ay =3a^2$

The equation(s) of the circle(s) which pass through the ends of the common chords of two circles

$2x^{2}+2y^{2}+8x+4y-7=0$ and

$x^{2}+y^{2}-8x-4y-5=0$ and touch the line

$x=7$ is (are) :

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$x^{2}+y^{2}-6x+2y-\dfrac{19}{4}=0$
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$x^{2}+y^{2}+120x+60y+11=0$
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$x^{2}+y^{2}-6x+2y+\dfrac{19}{4}=0$
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$x^{2}+y^{2}+120x+60y-11=0$

The equation of the circle which touches the axes of

$y$ at the origin and passing through

$(3,4)$ is

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$4(x^{2}+y^{2})-25x=0$
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$3(x^{2}+y^{2})-25x=0$
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$2(x^{2}+y^{2})-3x=0$
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$4(x^{2}+y^{2})-25x=0$

The value of k, such that the equation

$2x^{2}+2y^{2}-6x+8y+k=0$ represent a point circle , is

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0
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25
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$\frac {25}{2}$
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$-\frac {25}{2}$

Explanation

The given equation of circle is

$\Rightarrow$

$2x^2+2y^2-6x+8y+k=0$

$\Rightarrow$

$x^2+y^2-3x+4y+\dfrac{k}{2}=0$ [ Dividing both sides by

$2$ ]

Comparing it with general equation of circle

$x^2+y^2+2gx+2fy+c=0,$ we get

$\Rightarrow$

$2g=-3,\,2f=4$ and

$c=\dfrac{k}{2}$

$\therefore$

$g=\dfrac{-3}{2}$ and

$f=2$

Center

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

We know,

Radius

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

We know that for point circle radius is

$0.$

$r=0$

$\Rightarrow$

$0=\sqrt{\left(\dfrac{3}{2}\right)^2+(-2)^2-\dfrac{k}{2}}$

$\Rightarrow$

$0=\sqrt{\dfrac{9}{4}+4-\dfrac{k}{2}}$

$\Rightarrow$

$0=\dfrac{9}{4}+4-\dfrac{k}{2}$

$\Rightarrow$

$\dfrac{k}{2}=\dfrac{9+16}{4}$

$\Rightarrow$

$\dfrac{k}{2}=\dfrac{25}{4}$

$\therefore$

$k=\dfrac{25}{2}$

$If$

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

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$a\,\,circle\,with\,centre\,(1,1)$
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$a\,\,parabola\,with\,\,vertex\,(1,2)$
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$a\,parabola\,with\,directrix\,x = \frac{3}{2}$
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$a\,parabola\,with\,directrix\,x = - \frac{1}{2}$

If the radius of the circle

$x^{2}+y^{2}-18x-12y+k=0$ be

$11$ then k=

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34
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4
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-4
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49

Explanation

The equation of circle is

$x^2+y^2-18x-12y+k=0$ .

Comparing it with

$x^2+y^2+2gx+2fy+c=0$ we get,

$2g=-18,\,2f=-12$ and

$c=k$

$\therefore$

$g=-9,\,f=-6$ and

$c=k$

Center

$=(-g,-f)=(9,6)$

Radius of a circle

$(r)=11$ [ Given ]

We know,

$\Rightarrow$

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

$\Rightarrow$

$11=\sqrt{(-9)^2+(-6)^2-k}$

$\Rightarrow$

$11=\sqrt{81+36-k}$

$\Rightarrow$

$11=\sqrt{117-k}$

$\Rightarrow$

$121=117-k$

$\Rightarrow$

$4=-k$

$\therefore$

$k=-4$

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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