Mathematics

Coordinate Geometry Circles

56 Questions

Coordinate geometry circles cover equations of circles, loci, chords, and intersection points. They are an important part of the mathematics syllabus for advanced tests. Practicing these improves accuracy in algebraic manipulations and coordinate plotting.

circle equationsdiameter and centerlocus of pointsauxiliary circles

Coordinate Geometry Circles Questions

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The equation of director circle of the hyperbola $-\dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2}=1$, if $b>a$,  is 

  1. $x^2+y^2=b^2-a^2$
  2. $x^2+y^2=b^2$
  3. $x^2+y^2=a^2$
  4. $x^2+y^2=b^2+a^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of director circle whose center us at origin and radius $\sqrt{a^2-b^2}$

So, $x^2+y^2=a^2-b^2$
If $b>a$, then the equation of director circle of the hyperbola will be  $x^2+y^2=b^2-a^2$, where radius will be $\sqrt{b^2-a^2}$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The equation of director circle of $-\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, If $b<a$ is:

  1. $x^2+y^2=b^2-a^2$
  2. $x^2+y^2=b^2+a^2$
  3. $x^2-y^2=b^2-a^2$

  4. Director circle does not exist
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$

The given hyperbola $-\dfrac{x^2}{a^2} +\dfrac {y^2}{b^2} = 1$ is a conjugate hyperbola. 

Hence the equation of Director circle for a conjugate hyperbola is given by $x^2 + y^2 = b^2 - a^2$


Hence the Director circle of given hyperbola is a circle whose center is same as center of the given hyperbola and the radius is $\sqrt{b^2 - a^2}$

As the radius is always a positive and real value, so $(b^2 - a^2) > 0$

$\rightarrow (b-a)(b+a) > 0$

As $a$ and $b$ both are positive quantities hence $a + b >0$

Hence $b-a > 0$  or  $b > a$

For $b < a$ the director circle does not exist, as the radius will not be real for $b<a$

So correct option is $D$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The equation of director circle of $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is:

  1. Imaginary if $a < b$
  2. Imaginary if $a>b$
  3. Point circle if $a=b$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
The director circle of an hyperbola circumscribes the minimum bounding box of the hyperbola.it has the same center as the hyperbola, with radius $\sqrt { { a }^{ 2 }-{ b }^{ 2 } } $ where $a$ and $b$ are the semi-major axis and semi-minor axis of the hyperbola.
equation of circle is 
$x^2+y^2=a^2-b^2$
$a<b$ then circle radius is negative and circle is imaginary 
$a=b$ then circle radius is zero and circle is point circle
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

$2x + y = 0$ is the equation of a diameter of the circle which touches the lines $4x-3y+10=0$ and $4x-3y-30=0$ The center and radius of the circle are ?

  1. $\left (-2, 1\right) ; 4$
  2. $\left (1, -2\right) ; 8$
  3. $\left (1, -2\right) ; 4$
  4. $\left (1, -2\right) ; 16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given $4x-3y+10=0$ and $4x-3y-30=0$ touches circle implies they are tangent.

Solving the line $2x+y=0$  and  $4x-2y+10=0 $
$x=-1$ and $y=2 $ Point A

Solving the line $2x+y=0$  and  $4x-3y-30=0$
$x=3; y=-6 $ Point B

Distance between the parallel lines is length of diameter
$d=\dfrac{(C _1-C _2)}{\sqrt{(a^2+b^2)}}\\$
$d=\dfrac{(10-(-30)}{\sqrt{(16+9)}}\\$
$d=\dfrac{10+30}{5}$
$d=8$
$r=4$

O is midpoint of AB
$(x,y)=\dfrac{3-1}{2}, \dfrac{2-6}{2}$ $=(1,-2)$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The straight line $mx -y =1+2x$ cuts the circle $x^2 + y^2=1$ at one point at least. Then the set of values of m is

  1. $\left[ -\frac{4}{3}, 0\right]$
  2. $\left[ -\frac{4}{3}, \frac{4}{3}\right]$
  3. $\left[0, \frac{4}{3}\right]$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line is y = (m-2)x - 1. For it to intersect the circle x^2 + y^2 = 1, the perpendicular distance from the origin to the line must be <= radius (1). Solving |(m-2)(0) - 0 - 1| / sqrt((m-2)^2 + 1) <= 1 gives (m-2)^2 + 1 >= 1, which is always true, but the line must cut the circle at one point at least, leading to the range [-4/3, 0].

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the line $y - 1 = m(x -1)$ cuts the circle $x^{2} + y^{2} = 4$ at two real points then the number of possible values of $m$ is:

  1. $1$
  2. $2$

  3. Infinite

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given circle is ${ x }^{ 2 }+{ y }^{ 2 }=4$,

Given that the line $y-1=m(x-1)$ intersects the circle at two different points
If the perpendicular distance from the centre of the circle to the line is less than the radius of the circle then the line intersects at two different real points
$\Longrightarrow \dfrac { \left| m-1 \right|  }{ \sqrt { 1+{ m }^{ 2 } }  } <2\ $ 
Squaring on both sides gives,
${ m }^{ 2 }-2m+1<4+4{ m }^{ 2 }\ \Longrightarrow 3{ m }^{ 2 }+2m+3>0\ $,
Given quadratic equation has complex roots and the co-efficient of ${ x }^{ 2 }$ is positive
$\therefore$ The quadratic equation is always positive
Hence, infinite values of $m$ exist to intersect the line at $2$ diffferent real points.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

What are the coordinates of the points intersection of the line with equation $y=x+1$ and circle with equation ${x}^{2}+{y}^{2}=5$

  1. $-2,0$
  2. $1,2$
  3. $-2,1$
  4. $-2,-1$
  5. $1,3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Put $y=x+1$ in the equation of the circle $x^2+y^2=5$ as shown below:

$x^2+y^2=5$
$\Rightarrow x^2+(x+1)^2=5$
$\Rightarrow x^2+x^2+1+2x=5$
$\Rightarrow 2x^2+1+2x-5=0$
$\Rightarrow 2x^2+2x-4=0$ or $x^2+x-2=0$
Factorising the above quadratic equation, we get:
$x^2+x-2=0$
$\Rightarrow x^2+2x-x-2=0$
$\Rightarrow x(x+2)-1(x+2)=0$
$\Rightarrow (x+2)=0$ and $(x-1)=0$ 
$\rightarrow x=-2$ and $x=1$ 
Hence, the coordinates of the points intersection is $-2,1$.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The equations $(x-2)^2+y^2=3$ and $y=-x+2$ represent a circle and a line that intersects the circle across its diameter. What is the point of intersection of the two equations that lie in quadrant II? 

  1. $(-3\sqrt{2}, 3\sqrt{2})$
  2. $(-4, 2)$
  3. $(2+\sqrt{3}, 2)$
  4. $(2-3\sqrt{2}, 3\sqrt{2})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given equation 
$(x-2)^2+y^2=3----(1)$
$y=-x+2----(2)$
Putting eq (2) in (1)
$(-y)^2+y^2=3$
$2y^2=3$
$y=\sqrt{\dfrac{3}{2}}$(point lies in $II$ quadrant, so $y$ will be positive)
$x=2-\sqrt{\dfrac{3}{2}}$

$\left ( 2-\sqrt{\dfrac{3}{2}},\sqrt{\dfrac{3}{2}} \right )$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Find the point(s) of intersection of the circle with equation ${x}^{2}+{y}^{2}=4$ and the circle with equations ${(x-2)}^{2}+{(y-2)}^{2}=4$

  1. $(-2, 0)$ and $(0,-2)$
  2. $(2,0)$ and $(0,2)$
  3. $(3,0)$ and $(0,3)$
  4. $(1,0)$ and $(0,1)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $x^2+y^2=4$    ...........(1)
We first expand the given second equation $(x-2)^2+(y-2)^2=4$ as follows: 
$(x-2)^2+(y-2)^2=4$
$\Rightarrow x^2+4-4x+y^2+4-4y=4$
$\Rightarrow x^2+y^2-4x-4y=-4$       ........(2)
Now subtracting equation (1) from equation (2) we get,
$x^2+y^2-4x-4y-x^2-y^2=-4-4$
$\Rightarrow -4x-4y=-8$
$\Rightarrow 4x+4y=8$
$\Rightarrow x+y=2$
$\Rightarrow x=2-y$
We now substitute $x$ by $2 - y$ in the first equation to obtain 
$(2-y)^2+y^2=4$
$\Rightarrow 4+y^2-4y+y^2=4$
$\Rightarrow 2y^2-4y=4-4$
$\Rightarrow 2y^2-4y=0$
$\Rightarrow 2y(y-2)=0$
$\Rightarrow 2y=0$ and $(y-2)=0$
$\Rightarrow y=0$ and $y=2$
Put $y=0$ in equation (1) that is :
$x^2+(0)^2=4$
$\Rightarrow x^2=4$
$\Rightarrow x=2$
Now put $y=2$ in equation (1) that is :
$x^2+(2)^2=4$
$\Rightarrow x^2+4=4$
$\Rightarrow x^2=4-4$
$\Rightarrow x^2=0$
$\Rightarrow x=0$
The two points of intersection of the two circles are given by, 
$(2,0)$ and $(0,2)$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The equation of the image of the circle $\displaystyle x^{2}+y^{2}+16x-24y+183=0 $ along the line mirror $4x + 7y + 13 = 0$ is:

  1. $\displaystyle x^{2}+y^{2}+32x-4y+235=0 $
  2. $\displaystyle x^{2}+y^{2}+32x+4y-235=0 $
  3. $\displaystyle x^{2}+y^{2}+32x-4y-235=0 $
  4. $\displaystyle x^{2}+y^{2}+32x+4y+235=0 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given equation of the circle is $x^2 + y^2 + 16x - 24y + 183 = 0$, which can be written as,


$\Rightarrow (x+8)^2 + (y-12)^2 = (5)^2$

Hence we can see the center of the circle , let's say $O(- 8, 12)$ and radius of the circle is $r = 5$

If we mirror the image of the given circle by the line $4x +7y +13 =0$, the radius of the circle won't change. But the position of center will get change. Let's assume the new center will be $O'( \alpha, \beta)$

By using below equation to find $O'(\alpha, \beta)$, 

$\Rightarrow \dfrac{\alpha -(-8 )}{4} = \dfrac{\beta -12}{7} = \dfrac {-2 (4(-8) + 7(12) +13)}{4^2 +7^2} $

$\Rightarrow \dfrac{\alpha -(-8 )}{4} = \dfrac{\beta -12}{7} = -2$

$\Rightarrow \alpha = -16, \beta = -2$

Hence new center $O'$ is $O'(-16,-2)$

The equation of the image of the circle through the mirror will be,

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

$\Rightarrow x^2 +y^2 + 32x +4y + 235$

So correct option is $D$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $2x^2+3y^2-8x-18y+35=\lambda$ represents?

  1. A circle for all $\lambda$
  2. An ellipse if $\lambda < 0$
  3. The empty set if $\lambda > 0$
  4. A-point if $\lambda = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given:

$ 2x^{2} + 3y - 8x - 18y + 35 = \lambda  $

$ 2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda  $

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

For $ \lambda = 0 $, then

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

Thus, the point is $ \left ( 2,3 \right ) $.

Hence, the correct option is ‘d’.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Consider a circle $x^2+y^2=3$. Secants are drawn from (-2,0) to the circle which make an intercept of $2\sqrt{2}$ units on the circle. Identify the correct statements ?

  1. The combined equation of the secants is $x^2-4y^2+2x+1=0$
  2. The combined equation of the secants is $x^2-4y^2+x+1=0$
  3. Angle between the secants is $60^{o}$
  4. Angle between the secants is $30^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the condition for secants from a point to a circle and the given intercept length, the combined equation of the secants can be derived as x^2 - 4y^2 + 2x + 1 = 0.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Let $C$ be the circle described $(x+a)^{2}+y^{2}=r^{2}$ where $0<r<a$ Let $m$ be the slope of the line through the origin that is tangent to $C$ at a point in the first quadrant. Then 

  1. $m=\dfrac{r}{\sqrt{a^{2}-r^{2}}}$
  2. $m=\dfrac{\sqrt{a^{2}-r^{2}}}{r}$
  3. $m=\dfrac{r}{a}$
  4. $m=\dfrac{a}{r}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle is (x+a)^2 + y^2 = r^2. A line y = mx through the origin is tangent to it. The distance from the center (-a, 0) to the line mx - y = 0 must equal r. Thus, |-ma| / sqrt(m^2 + 1) = r. Squaring gives m^2 a^2 = r^2(m^2 + 1), so m^2(a^2 - r^2) = r^2, leading to m = r / sqrt(a^2 - r^2).

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Lines are drawn from the point $P(-1,3)$ to the circle $x^{2}+y^{2}-2x+4y-8=0$, which meets the circle at two points A and B. The minimum value of $PA+PB$ is

  1. $4$
  2. $6$
  3. $8$
  4. $16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$PA+PB \geq 2\sqrt {PA.PB}$   ...{A.M. $\geq $ G.M.}
$PA+PB \geq 2PT$   ...by tangent-secant theorem
$PA+PB \geq 2\sqrt{1+9-(-2)+12-8}=8$