Mathematics

Coordinate Geometry Circles

61 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 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
$PA+PB \geq 2\sqrt{1+9-(-2)+12-8}=8$

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

The locus of the centre of a circle touching the lines $x+2y=0$ and $x-2y=0$ is

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

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

let $(h,k)$ be the center
Then distance from center to both lines will be equal
$\left| \dfrac { h+2k }{ \sqrt { 5 }  }  \right| =\left| \dfrac { h-2k }{ \sqrt { 5 }  }  \right| $
$\Rightarrow hk=0$

Ans: A

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

The circle ${ x }^{ 2 }+{ y }^{ 2 }=4$ cuts the line joining the points $A(1,0)$ and $B(3,4)$ in two points P and Q. Let $\dfrac { BP }{ PA } =\alpha$ and $\dfrac { BQ }{ QA } =\beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation

  1. $3{ x }^{ 2 }+2x-21=0$
  2. $3{ x }^{ 2 }+2x+21=0$
  3. $2{ x }^{ 2 }+3x-21=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The equation of line joining $A(1,0)$ and $B(3,4)$ is $\dfrac{y-0}{x-1}=\dfrac{4-0}{3-1}\implies y=2x-2$    ... (1)

The point of intersection of this line and circle are

$x^2+(2x-2)^2=4\implies x^2+4x^2+4-8x=4\implies x=0,\dfrac{8}{5}$

Hence, points of intersection are $P(0,-2)$ and $Q\left(\dfrac{8}{5},\dfrac{6}{5}\right)$

Now, $BP=\sqrt{3^2+6^2}=\sqrt{45}$, $PA=\sqrt{1^2+2^2}=\sqrt{5}$, 

$BQ=\sqrt{\left(3-\dfrac{8}{5}\right)^2+\left(4-\dfrac{6}{5}\right)^2}=\sqrt{\dfrac{245}{25}}$ and $QA=\sqrt{\left(1-\dfrac{8}{5}\right)^2+\left(0-\dfrac{6}{5}\right)^2}=\sqrt{\dfrac{45}{25}}$

$\therefore \dfrac{BP}{PA}=3=\alpha$ and $\dfrac{BQ}{QA}=\dfrac{7}{3}=\beta$

The equation with roots $\alpha$ and $\beta$ is $(x-\alpha)(x-\beta)=0\implies (x-3)(3x-7)=0\implies 3x^2-16x+21=0$

This is the required answer.
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

If the area of the quadrilateral formed by the tangent from the origin to the circle $x^{2} +y^{2} +6x -10y

+ c = 0$ and the pair of radii at the points of contact of these tangents to tbe circle is $8$ square units. then $c$ is a root of the equation

  1. $ c^{2} -32c + 64 = 0$.
  2. $ c^{2} -34c + 64= 0$.
  3. $c^{2}+ 2c -64 = 0 $.
  4. $ c^{2} + 34c -64 = 0$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $OA, OB$ be the tangents from the origin to the given circle with centre $C(-3, 5)$
and radius .$\sqrt{9 + 25 -c}= \sqrt{ 34 -c} $
Then area of the quadribiteral $ OACB = 2 \times $ area of  $\triangle OAC = 2 \times (\dfrac 12) \times OA\times AC $
Now $OA =$ length of the tangent from the origin to the given circle $ = .\sqrt{C}$
and $AC =$ radius of the circle $=\sqrt{ 34 -c} $ so that. $\sqrt{C} \sqrt{34 -c} =8 $       ...(given)
$\Rightarrow  c (34 -c) =64 \Rightarrow c^{2} -34c+64=0$

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

The radius of the circle touching the straight lines $x-2y-1=0$ and $3x-6y+7=0$ is

  1. $\cfrac { 3 }{ \sqrt { 5 } } $
  2. $\cfrac { \sqrt { 5 } }{ 3 } $
  3. $\sqrt { 5 } $
  4. $\cfrac { 1 }{ \sqrt { 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Diameter of circle=distance of the point (1,0)
from $3x-6y+7=0$
$\therefore$ $\cfrac { 3(1)-6(0)+7 }{ \sqrt { { \left( 3 \right)  }^{ 2 }+{ \left( -6 \right)  }^{ 2 } }  } =\cfrac { 10 }{ \sqrt { 45 }  } =\cfrac { 2 }{ 3 } \sqrt { 5 } $
Now, radius of circle $=\cfrac { 1 }{ 2 } \left( \cfrac { 2 }{ 3 } \sqrt { 5 }  \right) =\cfrac { \sqrt { 5 }  }{ 3 } $

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

If the line $\displaystyle ax+by + c =0$ touches the circle $\displaystyle x^2 + y^2 -2x = \frac{3}{5}$ and is normal to the circle $\displaystyle x^2 + y^2 + 2x - 4y + 1 =0$, then $(a,b)$ are

  1. $(1, 3)$
  2. $(3, 1)$
  3. $(1, 2)$
  4. $(2, 1)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x^2+y^2-2x=\dfrac {3}{5}\Rightarrow (x-1)^2+y^2=\dfrac {8}{5}$

So, Radius, $R=2\sqrt {\dfrac {2}{5}}$ and it's center is at $(1,0)$

ie, Distance, $d$ from the circle to $ax+by+c=0$ is,
$d=\dfrac {a\times 1+b\times 0+c}{\sqrt{a^2+b^2}}=\dfrac {a+c}{\sqrt{a^2+b^2}} =2\sqrt {\dfrac {2}{5}}\longrightarrow (1)$ (Inorder to satisfy the criterion of a tangent)

$x^2+y^2+2x-4y+1=0 \Rightarrow (x+1)^2+(y-2)^2=4$
So, It's center is at $((-1),2)$
As $ax+by+c=0$ is normal to the circle, it should go through the centre of the circle.
ie, $a-2b=c$ and $(y-2)=m(x+1)\longrightarrow (2)$

Substituting $c$ in (1),
$\dfrac {a+(a-2b)}{\sqrt{a^2+b^2}} =2\sqrt {\dfrac {2}{5}}$
$\Rightarrow \dfrac {a-b}{\sqrt {a^2+b^2}}=\sqrt {\dfrac {2}{5}}$

So, we can say $(a-b)=k\sqrt {2}$ and $a^2+b^2=5k^2$ foe some constant $k$.
$a^2+b^2-(a-b)^2=2ab=5k^2-2k^2=3k^2$
$(a-b)^2+4ab=(a+b)^2=6k^2+2k^2=8k^2\Rightarrow (a+b)=2k\sqrt{2}$
$a=\dfrac {1}{2}((a+b)+(a-b))=\dfrac {1}{2}(3k\sqrt{2})$
$b=\dfrac {1}{2}((a+b)-(a-b))=\dfrac {1}{2}(k\sqrt {2})$

Slope of the line, $m=\dfrac {dy}{dx}$
$\dfrac {d}{dx}(ax+by+c)=0\Rightarrow a+b\dfrac {dy}{dx}=0$
ie, $m=\dfrac {(-a)}{b}=(-3)$ (from above equations of $a$ and $b$)

Substituting the slope in (2),
$(y-2)=(-3)(x+1)\Rightarrow 3x+y+1=0$

Compairing with general equation given,
$(a,b)=(3,1)$

Option B is the correct answer.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The equation $x^2y^2 - 9y^2- 6x^2 y + 54y = 0$ represents

  1. A pair of straight lines and a circle

  2. A pair of straight lines and a parabola

  3. A set of four straight lines forming a square

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

$x^2y^2-9y^2-6x^2y+54y=0$
$y^2(x^2-9)-6y(x^2-9)=0$
$(y^2-6y)(x^2-9)=0$
$y(y-6)(x+3)(x-3)=0$
Therefore the lines are
$y=0$
$y=6$
$x=3$
$x=-3$
The above set of lines represent a square of side $6$ units.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

If $(6, -3)$ is the one extremity of diameter to the circle $x^{2}+y^{2}-3x+8y-4=0$ then its other extremity is-

  1. $(3/2, -4)$
  2. $(-3, -5)$
  3. $(3, -5)$
  4. $(3, 5)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The center of the circle x^2 + y^2 - 3x + 8y - 4 = 0 is (3/2, -4). Since the center is the midpoint of the diameter, if one end is (6, -3), let the other be (x, y). Then (6+x)/2 = 3/2 implies x = -3, and (-3+y)/2 = -4 implies y = -5.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

The coordinates of the centre of a circle are $(-6,1.5)$. If the ends of a diameter are $(-3,y)$ and $(x, -2)$ then:

  1. $x= 9, y=5$
  2. $x=5, y= -9$
  3. $x=-9, y=5$
  4. $x=-9, y=-5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The centre of the circle lies at the mid point of the diameter.
Mid-point of two points $ { (x } _{ 1 },{ y } _{ 1 }) $ and $ { (x } _{ 2 },{ y } _{
2 }) $ is  calculated by the formula $ \left( \dfrac { { x } _{ 1 }+{ x
} _{ 2 } }{ 2 } ,\dfrac { { y } _{ 1 }+y _{ 2 } }{ 2 }  \right) $
Using this formula, mid point of $ (3,y) $ and $ (x,2) = \left( \dfrac { -3 + x }{ 2 } ,\dfrac { y - 2 }{ 2 } \right) $
So,$\left( \dfrac { -3 + x }{ 2 } ,\dfrac { y - 2 }{ 2 }  \right) = (-6, 1.5) $
$ \Rightarrow  - 3 + x = -6 \times 2 $ and $ y - 2 = 1.5 \times 2 $
$\Rightarrow x = - 9 ; y = 5 $

Multiple choice position of point wrt ellipse ellipse maths

Equation of the largest circle with centre (1,0) that can be inscribed in the ellipse $x^2 + 4y^2 = 16$ is 

  1. $2x^2 + 2y^2 - 4x + 7 = 0$
  2. $x^2 + y^2 - 2x + 5 = 0$
  3. $3x^2 + 3y^2 - 6x - 8 = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{4}=1$

Point on the ellipse $(4\cos\theta, 2\sin\theta)$

Let the circle have radius $=r$

$(x-1)^{2}+(y-0)^{2}=r^{2}$ Solving if with ellipse

$x^{2}+4y^{2}=16$

$(x-1)^{2}+\dfrac{(16-x^{2})}{4}=r^{2}$

$4(x^{2}-2x+1)+16-x^{2}=4r^{2}$

$3x^{2}-8x+20-4r^{2}=0$

As the circle & ellipse touch each other 

$D=0$

$8^{2}-4.2\times (20-4r^{2})=0$

$r^{2}=\dfrac{\pi}{3}$

$(x-1)^{2}+y^{2}=\dfrac{11}{3}$

$3x^{2}+3y^{2}-6x-8=0$
Multiple choice

What is the equation of the circle with center ((2, 3)) and radius 5?

  1. \((x - 2)^2 + (y - 3)^2 = 25\)
  2. \((x - 2)^2 + (y - 3)^2 = 10\)
  3. \((x - 2)^2 + (y - 3)^2 = 15\)
  4. \((x - 2)^2 + (y - 3)^2 = 20\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of a circle with center ((h, k)) and radius r is given by the formula ((x - h)^2 + (y - k)^2 = r^2). Substituting h = 2, k = 3, and r = 5 into the formula, we get: ((x - 2)^2 + (y - 3)^2 = 5^2) = ((x - 2)^2 + (y - 3)^2 = 25). Therefore, the equation of the circle with center ((2, 3)) and radius 5 is ((x - 2)^2 + (y - 3)^2 = 25).

Multiple choice

What is the equation of a circle with center (h, k) and radius r?

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

  2. (x + h)^2 + (y + k)^2 = r^2

  3. (x - h)^2 - (y - k)^2 = r^2

  4. (x + h)^2 - (y + k)^2 = r^2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of a circle with center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = r^2.