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 general knowledge math & puzzles

What is the equation of a circle of radius 6 units centered at (3, 2)?

  1. x2 + y2 + 6x – 4y = 23

  2. x2 + y2 - 6x – 4y = 23

  3. x2 + y2 + 6x + 4y = 23

  4. x2 + y2 - 6x – 4y = - 23
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of a circle is (x-h)^2 + (y-k)^2 = r^2. For center (3,2) and radius 6, this is (x-3)^2 + (y-2)^2 = 36. Expanding gives x^2 - 6x + 9 + y^2 - 4y + 4 = 36, which simplifies to x^2 + y^2 - 6x - 4y = 23.

Multiple choice intersection of a line and a parabola conic section maths

Let the equation of a circle and a parabola be $x^2+y^2-4x-6=0$ and $y^2=9x$ respectively. Then

  1. $\left ( 1,-1 \right )$ is a point on the common chord of contact
  2. the equation of the common chord is $y+1=0$
  3. the length of the common chord is $6$

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

Given parabola and circle are $y^2=9x$ and $x^2+y^2-4x-6=0$ respectively.
Now solving these, $x^2+9x-4x-6=0\Rightarrow x^2+5x-6=0\Rightarrow (x-1)(x+6)=0\Rightarrow x = 1,-6$ but $x < 0$ is not a solution
Thus $x = 1, $ and corresponding $y = 3,-3$
Hence point of intersection of the parabola and circle are $(1,-3), (1,3)$
Hence equation of common chord is given by, $x=1$
and length of common chord $=3-(-3)=6$

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

A square is inscribed in the circle $x^2 + y^2 -2x +4y - 93 = 0$ with its sides parallel to the coordinates axes. The coordinates of its vertices are 

  1. $( - 6, - 9), \, ( - 6, 5), \, (8, - 9)$ and $(8, 5)$
  2. $( - 6, 9), \, ( - 6, - 5), \, (8, - 9)$ and $(8, 5)$
  3. $( - 6, - 9), \, ( - 6, 5), \, (8, 9)$ and $(8, 5)$
  4. $( - 6, - 9), \, ( - 6, 5), \, (8, - 9)$ and $(8, - 5)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle equation is (x-1)^2 + (y+2)^2 = 93 + 1 + 4 = 98. The radius is sqrt(98) = 7*sqrt(2). For a square with sides parallel to axes, the distance from center (1, -2) to vertices is the radius.

Multiple choice maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

A circle with centre $(1,1)$ intersects X axis at $(1,0)$ and  Y axis at $(0,1)$. Find the centre of the circle when reflected through X axis.

  1. $(1,-1)$
  2. $(1,1)$
  3. $(-1,-1)$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The graph is symmetric(reflection) w.r.t to the $x$-axis i.e. $y=0$ , If $(x,y)$ is a point on the graph ,then $(x,-y)$ is also a point on the graph .

Given center at $(1,1)$
using above concept reflection of center about $X$-axis lies at  $(1,-1)$
Multiple choice decimal representation of rational numbers rational and irrational numbers maths

The point $\left( \sin { \theta  } ,\cos { \theta  }  \right) ,\theta $ being any real number, lie inside the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+\lambda =0$, if

  1. $\lambda <1+2\sqrt { 2 } $
  2. $\lambda >2\sqrt { 2 } -1$
  3. $\lambda <-1-2\sqrt { 2 } $
  4. $\lambda >1+2\sqrt { 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+\lambda =0$
Radius of circle$=\sqrt { 1+1-\lambda  } $
$=\sqrt { 2-\lambda  } $
Maximum distance of $\left( \sin { \theta  } ,\cos { \theta  }  \right) $
From center of above circle is when $\theta =\cfrac { 5\pi  }{ 4 } $
Thus distance$=\sqrt { { \left( 1+\cfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 }+{ \left( 1+\cfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 } } $\
$=\left( \sqrt { 2 } +1 \right) $
$\therefore \sqrt { 2-\lambda  } >\sqrt { 2+1 } $ [For all points to lie in center]
$\therefore 2-\lambda >2+1+2\sqrt { 2 } $
$\lambda <2-3-2\sqrt { 2 } $
$\lambda <-1-2\sqrt { 2 } $
Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

Circle on which the coordinates of any point are $(2+4 \cos \theta,-1+4 \sin \theta)$ where $\theta$ is the parameter is given by $(x-2)^2+(y+1)^2=16$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
From given conditions, we have,
$x=2+4 \cos \theta $
$x-2=4\cos \theta$           .......(1)
$y=-1+4\sin \theta$
$y+1=4 \sin \theta$           .......(2)
Squaring and adding equation 1 and 2, we get,
$(x-2)^2+(y+1)^2=16$.
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The abscissa of two points A and B are the roots of the equation ${x^2} + 2ax - {b^2}$ and their ordinates are the root of the equation ${x^2} + 2px - {q^2}=0$. the equation of the circle with AB as diameter is 

  1. ${x^2} + {y^2} + 2ax + 2py + {b^2} + {q^2} = 0$
  2. ${x^2} + {y^2} - 2ax - 2py - {b^2} - {q^2} = 0$
  3. ${x^2} + {y^2} + 2ax + 2py - {b^2} - {q^2} = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
${ x }^{ 2 }+2ax-{ b }^{ 2 }=0$

Let roots be ${ x } _{ 1 }$ & ${ x } _{ 2 }$

${ x } _{ 1 }+{ x } _{ 2 }=\dfrac{-b}{a}=-2a$

$ { x } _{ 1 }{ x } _{ 2 }=\dfrac{c}{a}=-{ b }^{ 2 }$

$ { x }^{ 2 }+2px-{ q }^{ 2 }=0$

Let roots be ${ y } _{ 1 }$ & ${y } _{ 2 }$

${ y } _{ 1 }+{ y } _{ 2 }=-2p$

$ { y} _{ 1 }{ y } _{ 2 }=-{ q }^{ 2 }$

Equation in diametric form is 
${ x }^{ 2 }+{ y }^{ 2 }-({ x } _{ 1 }+{ x } _{ 2 })x-({ y } _{ 1 }+{ y } _{ 2 })y+{ x } _{ 1 }{ x } _{ 2 }+{ y } _{ 1 }{ y } _{ 2 }=0$

$ { x }^{ 2 }+{ y }^{ 2 }+2ax+2py-{ b }^{ 2 }-{ q }^{ 2 }=0$
Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

An equation of sphere with centre at origin and radius $r$ can be represented as

  1. $x^2+y^2+z^2=r$
  2. $x^2+y^2+z^2=r^2$
  3. $x^2+y^2+z^2=2r^2$

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

Sphere is locus of a point in 3D whose distance from a fixed point(center) is constant (radius)

$\Rightarrow \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=|r|$
$\Rightarrow x^2+y^2+z^2=r^2$, square both sides 

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation $\frac{x^2}{1-k}-\frac{y^2}{1+k}=1$, $k<1$ represents 

  1. $circle$
  2. $ellipse$
  3. $hyperbola$
  4. $none$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Equating the above equation with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{1-k}-\dfrac{{y}^{2}}{1+k}=1$
we get $A=\dfrac{1}{1-k}, B=0, C=\dfrac{1}{1+k},D=0,E=0$ and $F=-1$
$(i)$For the second degree equation to represent a circle , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC=0$ and also $A=C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}=0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}=0$
This case does not exist
$(ii)$For the second degree equation to represent a ellipse , the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}<0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}>0$
$\Rightarrow 1-{k}^{2}<0$
$\Rightarrow -{k}^{2}<-1$
$\Rightarrow {k}^{2}>1$ does not exist since it is given that $k<1$
$(iii)$For the second degree equation to represent a hyperbola, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC>0$ and also $A\neq C$
$\Rightarrow -4\times \dfrac{1}{1-k}\times \dfrac{1}{1+k}>0$
$\Rightarrow \dfrac{1}{1-{k}^{2}}<0$
$\Rightarrow 1-{k}^{2}>0$
$\Rightarrow -{k}^{2}>-1$
$\Rightarrow {k}^{2}<1$ 
$\therefore k<1$
Hence the above equation represents a hyperbola.
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

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

  1. a circle

  2. hyperbola

  3. a pair of lines

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

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 $\dfrac{x^2}{64}-\dfrac{y^2}{49}=1$ is

  1. $x^2+y^2=15$
  2. $x^2+y^2=64$
  3. $x^2+y^2=18$
  4. $x^2+y^2=10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of hyperbola is $\dfrac { { x }^{ 2 } }{ 64 } -\dfrac { { y }^{ 2 } }{ 49 } =1$

Here $a=8,b=7$
Equation of director circle for hyperbola is
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }\ { x }^{ 2 }+{ y }^{ 2 }=64-49\ { x }^{ 2 }+{ y }^{ 2 }=15$
Hence, option A is correct.

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 for $\dfrac{x^2}{100}-\dfrac{y^2}{36}=1$, is

  1. $2x^2+2y^2=100$
  2. $\sqrt 2x^2+\sqrt 2y^2=100$
  3. $x^2+y^2=6$
  4. $x^2+y^2=64$
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 $x^2 + y^2 = a^2 - b^2$


Here the given hyperbola is $\dfrac{x^2}{100} -\dfrac {y^2}{36} = 1$,

Here $a =10$ and $b =6$

So equation of the director circle will be $x^2 + y^2 = (10)^2 - (6)^2$

$\Rightarrow x^2 + y^2 = 64$

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 hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is

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


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

So correct option is $D$.