Mathematics

Geometry and Trigonometry

115 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

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.

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 the origin to the circle $x^{2} + y^{2} - 2px - 2qy + q^{2} = 0$ are perpendicular if

  1. $p = q$
  2. $p^{2} = q^{2}$
  3. $q = -p$
  4. $p^{2} + q^{2} = 1$.
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Equation of the given circle can be written as $(x -p)^{2} + (y -q)^{2} = p^{2}$
so, that the centre of the circle is $(p, q)$ and its radius is $p$.
This shows that $x = 0$ is a tangent to the circle from the origin.
Since tangents from the origin are perpendicular, the equation of the other tangent must be $y = 0$,
which is possible if $q = \pm  p $  or $p^{2} =q^{2}$

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

The angle between the two tangents from the origin to the circle ${(x-7)}^{2}+{(y+1)}^{2}=25$ equals-

  1. $\cfrac{\pi}{2}$
  2. $\cfrac{\pi}{3}$
  3. $\cfrac{\pi}{4}$

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

Center is $(7,-1)$ and radius $=5$
Let equation of tangent from the origin be $y=mx$ $\Rightarrow mx-y=0$
Then, $\displaystyle\left| \frac { 7m+1 }{ \sqrt { { m }^{ 2 }+1 }  }  \right| =5$
$\Rightarrow { \left( 7m+1 \right)  }^{ 2 }=25\left( { m }^{ 2 }+1 \right) \Rightarrow 24{ m }^{ 2 }+14m-24=0$
Let ${ m } _{ 1 }$ and ${ m } _{ 2 }$ be the slopes of the two tangents.
Since $\displaystyle{ m } _{ 1 }{ m } _{ 2 }=-\frac{24}{24}=-1$
The two tangents are at right angles.

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 the origin to the circle ${ x }^{ 2 }+{ y }^{ 2 }-2rx-2hy+{h}^{2}=0$ are perpendicular if-

  1. $h=r$
  2. $h=-r$
  3. ${r}^{2}+{h}^{2}=1$
  4. ${r}^{2}+{h}^{2}=2$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Equation of the given circle can be written as ${ \left( x-r \right)  }^{ 2 }+{ \left( y-h \right)  }^{ 2 }={ p }^{ 2 }$
This has $(r,h)$ as the center and $r$ as the radius showing that it touches $y-$axis.
$\Rightarrow$ Other tangent from the origin to the circle must be $x-$axis which is possible if $h=\pm r$

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

If the tangents $PA$ and $PB$ are drawn from the point $P(-1,2)$ to the circle ${ x }^{ 2 }+{ y }^{ 2 }+x-2y-3=0$ and $C$ is the center of the circle, then the area of the quadrilateral $PACB$ is 

  1. $4$
  2. $16$

  3. Does not exists

  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given circle is $S:{ x }^{ 2 }+{ y }^{ 2 }+x-2y-3=0$

Since at point $P\left( -1,2 \right) $ ${ S } _{ \left( -1,2 \right)  }=1+4-1-4-3=-3<0$
the point $P(-1,2)$ lies inside the circle.
Consequently, the tangents from the point $P(-1,2)$ to the circle does not exits.
Thus, the quadrilateral $PACB$ cannot be formed.

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

In the given figure, if $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle APB=54^{\circ},$ then $\angle OAB$ equals

  1. $16^{\circ}$
  2. $18^{\circ}$
  3. $27^{\circ}$
  4. $36^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $PA$ and $PB$ are the tangents from the point P.
$\angle APB = 54^{\circ}$
Now, In quadrilateral AOBP
$\angle OAP = \angle OBP = 90^{\circ}$ (Angle between tangent and radius)
Sum of angles = 360
$\angle OAP + \angle OBP + \angle OAB + \angle APB = 360$
$90 + 90 + 54 + \angle AOB = 360$
$\angle AOB = 126$

Now, In $\triangle OAB$
$OA = OB$ (Radius of circle)
$\angle OAB = \angle OBA$ (Isosceles triangle property)
Sum of angles = 180
$\angle OAB + \angle OBA + \angle AOB = 180$
$2 \angle OAB + 126 = 180$
$\angle OAB = 27^{\circ}$