Mathematics

Geometry and Trigonometry

94 Questions

Geometry and trigonometry questions cover circles, tangents, and trigonometric ratios. Problems often combine algebraic geometry with angle properties to test spatial reasoning. This forms a core component of the mathematics section in engineering and civil services exams.

Circle tangentsTrigonometric ratiosHyperbola propertiesEllipse tangentsSecant construction

Geometry and Trigonometry Questions

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

Tangents $TP$ and $TQ$ are drawn from a point $T$ to circle $x^{2}+y^{2}=a^{2}$. If the point $T$ lies on the line $px+qy=r$, then locus of the centre of circumcircle of $\triangle TPQ$ is

  1. straight line

  2. circle

  3. parabola

  4. ellipse
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circumcircle of triangle TPQ has the segment OT as its diameter, where O is the center of the circle and T is the external point. If T lies on a line, the locus of the midpoint of OT will also be a straight line.

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

Tangents PA and PB are drawn to the cicle $S\, \equiv \,{x^2}\, + \,{y^2}\, - \,2y\, - \,3\, = \,0$ from the point $P(3, 4)$. Which of the following alternative(s) is/are correct ?

  1. The power of point $P(3, 4)$ with respect to circle $S=0$ is $14$.
  2. The angle between tangents from $P(3, 4)$ to the circle $S=0$ is $\frac{\pi }{3}$
  3. The equation of circumcircle of $\Delta PAB\,$ is ${x^2}\, + \,{y^2}\, - \,3x\, - \,5y\, + \,4\, = 0$
  4. The area of quadrilateral $PACB$ is $3\sqrt 7 $ square units where C is the centre of circle $S = 0$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The power of a point (x1, y1) with respect to a circle x^2 + y^2 + 2gx + 2fy + c = 0 is x1^2 + y1^2 + 2gx1 + 2fy1 + c. For P(3,4) and S = x^2 + y^2 - 2y - 3 = 0, Power = 3^2 + 4^2 - 2(4) - 3 = 9 + 16 - 8 - 3 = 14.

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

If $OA$ and $OB$ are the tangents to the circle ${x}^{2}+{y}^{2}-6x-8y+21=0$ drawn from the origin $O$, then $AB$ equals 

  1. ${ \dfrac { 17 }{ 3 } } $
  2. $\dfrac { 4 }{ 5 } \sqrt { 21 }$
  3. $11$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
equation of circle $\Rightarrow x^2+y^2-6x-8y+21=0$
radius $\Delta =\sqrt {9+16-21}=2$
$AB$ is chord of contact & its equation is
$x. x _1 +yy _1+9(x+x _1)+f(y+y _1)+c=0$
$(x _1, y _1)=(0,0)$
$0+0-3(x+0)-4(y+0)+21=0$
$3x+4y-21=0$
Perpendicular distance from $(3, 4)$ to line $l _1$
$CM=\dfrac {3(3)+4(4)-21}{\sqrt {9+16}}=\dfrac {4}{5}$
$AM=\sqrt {AC^2-CH^2}=\sqrt {4-\dfrac {16}{25}}=\dfrac {2}{5}\sqrt {21}$
$AB=2AM=\dfrac {4}{5}\sqrt {21} $ 
option $B$ is correct.


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

If 't$ _{1}$','t$ _{2}$','t$ _{3}$'are the lengths of the tangents drawnfrom centre of ex-circle to the circum circle of the $ \Delta A B C $, then- $ \frac { 1 } { t _ { 1 } ^ { 2 } } + \frac { 1 } { t _ { 2 } ^ { 2 } } + \frac { 1 } { t _ { 3 } ^ { 2 } } = $

  1. $ \frac { a b c } { a + b + c } $
  2. $ \frac { a b c } { a - b + c } $
  3. $ \frac { 2 a b c } { a + b + c } $

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

This is a known identity in triangle geometry relating the lengths of tangents from the excenter to the circumcircle.

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

$y=mx+b$ is a tangent to the circle ${x}^{2}+{y}^{2}-6x=16\ if\ \left (3\ m+b\right)^{2}=5\left (1+{m}^{2}\right)$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The condition for a line y = mx + b to be tangent to a circle x^2 + y^2 - 2gx - 2fy + c = 0 is (mg + f + b)^2 = r^2(1 + m^2). For x^2 + y^2 - 6x - 16 = 0, the center is (3, 0) and r^2 = 16 + 9 = 25. Substituting g=3, f=0, r^2=25 gives (3m + b)^2 = 25(1 + m^2), not 5(1+m^2).

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

The tangents drawn from origin to the circle ${ x }^{ 2 }+{ y }^{ 2 }-2ax-2by+{ b }^{ 2 }=0$ are perpendicular to each other, if

  1. $a-b=1$
  2. $a+b=1$
  3. ${ a }^{ 2 }-{ b }^{ 2 }=0$
  4. ${ a }^{ 2 }+{ b }^{ 2 }=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given circle equation ${x}^{2}+{y}^{2}-2ax-2by+{b}^{2}=0$
Center $: (a,b)$ and Radius $= \sqrt { {a}^{2}+{b}^{2}-{b}^{2} } $
Both tangents are drawn from origin and perpendicular to each other. So, two tangent are $x$ and $y$ axis.
Hence, $\sqrt { {a}^{2}+{b}^{2}-{b}^{2} } = a = b$
$\Rightarrow {a}^{2}={b}^{2}$
$\Rightarrow {a}^{2}-{b}^{2} = 0$ 

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

State whether the statement is true/false 

Two tangents $TP$ and $TQ$ are drawn to a circle with center $O$ from an external point $T$, then  $\angle PTQ=\angle OPQ$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a quadrilateral formed by the center O, the two points of tangency, and the external point T, the angles at the points of tangency are 90 degrees. Thus, angle PTQ + angle POQ = 180 degrees. The statement angle PTQ = angle OPQ is generally false.

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

If from a point P, two perpendicular tangents are drawn to the circle ${x^2} + {y^2} - 2x + 2y = 0$, then the coordinates of point P cannot be 

  1. $(3, - 1)$
  2. $(1,1)$
  3. $(\sqrt 3 + 1,0)$
  4. $(2,\sqrt 3 + 1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The locus of points from which perpendicular tangents are drawn to a circle is the director circle. For x^2 + y^2 - 2x + 2y = 0, the center is (1, -1) and r^2 = 1 + 1 = 2. The director circle is (x-1)^2 + (y+1)^2 = 2(2) = 4. Point (2, sqrt(3)+1) gives (2-1)^2 + (sqrt(3)+1+1)^2 = 1 + (sqrt(3)+2)^2 = 1 + 3 + 4 + 4sqrt(3) = 8 + 4sqrt(3), which is not 4.

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

The number of tangents to the circle ${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+9=0$ which passes through the point $(3,-2)$ is

  1. $2$
  2. $1$
  3. $0$

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

Let $S\equiv { x }^{ 2 }+{ y }^{ 2 }-8x-6y+9=0$

Now $s$ for $(-3,2)=9+4-24+12+9>0$
$\therefore$ the point $(3,-2)$ lies outside the circle.
$\therefore$ $2$ tangents can be drawn to the circle from the point $(3,-2)$

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

Tangents drawn from the origin to the circle $ \displaystyle x^{2}+y^{2}-2px-2qy+q^{2}=0 $ are perpendicular to each other if

  1. $ \displaystyle p^{2}=q^{2} $
  2. $ \displaystyle p^{2}-q^{2}= 1 $
  3. $ \displaystyle p^{2}+q^{2}= 1 $

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

The equation of pair of tangents drawn from the origin to the given circle are $S{ S } _{ 1 }={ T }^{ 2 }$
$\Rightarrow \left( { x }^{ 2 }+{ y }^{ 2 }-2px-2qy+{ g }^{ 2 } \right) \left( 0+0-0-0+{ g }^{ 2 } \right) ={ \left( x.0+y.0-p\left( x+0 \right) -q\left( y+0 \right) +{ y }^{ 2 } \right)  }^{ 2 }$
$\Rightarrow { q }^{ 2 }\left( { x }^{ 2 }+{ y }^{ 2 }-2px-2qy+{ g }^{ 2 } \right) -{ \left( -px-qy+{ g }^{ 2 } \right)  }^{ 2 }=0$
The two tangents are $\bot $ if ${ g }^{ 2 }+{ q }^{ 2 }-{ p }^{ 2 }-{ g }^{ 2 }=0$
(Sum of coefficient of ${ x }^{ 2 }+{ y }^{ 2 }=0$)
$\Rightarrow { q }^{ 2 }={ p }^{ 2 }$

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

If the distance from the origin of the centers of the three circles ${ x }^{ 2 }+{ y }^{ 2 }+2{ a } _{ i }x={ a }^{ 2 }\left( i=1,2,3 \right) $ are in G.P., then the length of the tangent drawn to them from any point on the circle ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ are in

  1. A.P.

  2. G.P.

  3. H.P.

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

The centers of the three given circles are $\left( -{ \alpha  } _{ 1 },0 \right) ,\left( -{ \alpha  } _{ 2 },0 \right) $ and $\left( -{ \alpha  } _{ 3 },0 \right) $.

the distance of the three points from the origin are ${ \alpha  } _{ 1 },{ \alpha  } _{ 2 }$ and ${ \alpha  } _{ 3 }$.
Given: ${ \alpha  } _{ 1 },{ \alpha  } _{ 2 }$ and ${ \alpha  } _{ 3 }$ are in G.P.
$\Rightarrow { { \alpha  } _{ 2 } }^{ 2 }={ \alpha  } _{ 1 }{ \alpha  } _{ 2 }$
Now, coordinate of any point on the circle ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ are $\left( a\cos { \theta  } ,a\sin { \theta  }  \right) $.
$\therefore$ The lengths of the tangents drawn from the point $\left( a\cos { \theta  } ,a\sin { \theta  }  \right) $ to the three given circles are
$\sqrt { 2{ \alpha  } _{ 1 }a\cos { \theta  }  } ,\sqrt { 2{ \alpha  } _{ 2 }a\cos { \theta  }  } $ and $\sqrt { 2{ \alpha  } _{ 3 }a\cos { \theta  }  } $
using (1) are in G.P.

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

Two $ \displaystyle \perp $ tangents to the circle $ \displaystyle x^{2}+y^{2}=a^{2} $ meet at a point P. The locus of P has the equation

  1. $ \displaystyle x^{2}+y^{2}=3a^{2} $
  2. $ \displaystyle x^{2}+y^{2}=2a^{2} $
  3. $ \displaystyle x^{2}+y^{2}=4a^{2} $

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

The coordinates of $P$ be $(h,k)$. Then the equation of the tangents drawn from $P(h,k)$ to ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ is
$\left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) \left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 } \right) ={ \left( hx+hy-{ a }^{ 2 } \right)  }^{ 2 }$   (using SS'$={ T }^{ 2 }$)
This equation represents a pair of perpendicular lines.
Therefore, coefficient of ${ x }^{ 2 }$$+$ coefficient of ${ y }^{ 2 }=0$
$\Rightarrow \left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 }-{ h }^{ 2 } \right) +\left( { h }^{ 2 }+{ k }^{ 2 }-{ a }^{ 2 }-{ k }^{ 2 } \right) =0$
$\Rightarrow { h }^{ 2 }+{ k }^{ 2 }={ 2a }^{ 2 }$
Hence, locus is ${ x }^{ 2 }+{ y }^{ 2 }=2{ a }^{ 2 }$

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

If the length of the tangent drawn from any point on the circle $\displaystyle x^{2}+y^{2}+15x-17y+c^{2}=0$ to the circle $\displaystyle x^{2}+y^{2}+15x-17y+21=0 \ is \ \sqrt{5}$ units , then $c$ is equal to

  1. $-3$
  2. $3$
  3. $-4$
  4. $4$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation
Required length
$\sqrt { { x }^{ 2 }+{ y }^{ 2 }+15x-17y+21-\left( { x }^{ 2 }+{ y }^{ 2 }+15x-17y+{ c }^{ 2 } \right)  } =\sqrt { 5 } $
$\Rightarrow \sqrt { 21-{ c }^{ 2 } } =\sqrt { 5 } \Rightarrow { c }^{ 2 }=16\Rightarrow c=\pm 4$
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A line is drawn through the point $P(3, 11)$ to cut the circle $x^{2}+y^{2}= 9$ at $A$ and $B$. Then $PA\cdot PB$ is equal to

  1. $9$
  2. $121$
  3. $ 205$
  4. $139$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From geometry we know $PA\cdot  PB = (PT)^{2}$

where $PT$ is the length of the tangent from $P$ to the circle.

Hence $PA\cdot PB=

(3)^2 + (11)^{2} - 9 = 11^{2} = 121$

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

If $t _{i}$ is the length of the tangent to the circle $ x^{2}+ y^{2} + 2g _{i} x + 5 =0; i =1,2,3$ from any point and $g _{1}, g _{2}$ and $g _{3} $ are in A.P. and $A _{i} = (g _{i},- t _{i}^{2})$, then

  1. $A _{1}, A _{2}, A _{3} $are collinear
  2. $A _{2}$ is the mid-point of $A _{1}$ and $A _{3} $
  3. $ A _{1} A _{2} $ is perpendicular. to $A _{2} A _{3}$
  4. $A _{2}$ divides $A _{1} A _{3}$ in the ratio $2: 5$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$t _{i}^{2} = x^{2} + y^{2} + 2g _{i}x + 5$  where $(x, y) $ is any point. 


Since  $g _{1}, g _{2}, g _{3}$ are in $A.P.$

$\Rightarrow 2g _{2} = g _{1} + g _{3}$
$\Rightarrow  2t _{2}^{2}  = t _{1}^{2} + t _{3}^{2} \Rightarrow  t _{1}^{2},t _{2}^{2} ,t _{3}^{2} $ are in $A.P.$
and $A _{2}$  is the mid-point of $A _{1}$ and $A _{3}$.

$\Rightarrow  A _{1}, A _{2}, A _{3}$  are collinear.