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

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
$\angle OAB + \angle OBA + \angle AOB = 180$
$2 \angle OAB + 126 = 180$
$\angle OAB = 27^{\circ}$

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 $\displaystyle \left ( x-7 \right )^{2}+\left ( y+1 \right )^{2}=25 $ equals

  1. $\displaystyle \frac{\pi }{4}$
  2. $\displaystyle \frac{\pi }{3}$
  3. $\displaystyle \frac{\pi }{2}$

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $y + 1 = m (x - 7) + \sqrt{25}(\sqrt{m^2 + 1})$ be any line to the circle.

Since we need tangents form $(0,0)$

$(0+1) = m(0 – 7) + 5\sqrt{m^2 + 1}$

$(7m + 1)^2 = 25(m^2 + 1)$

$\implies 24m^2 + 14m – 24 =0$

If $m _1, m _2$ are roots of the equation

$m _1m _2 = \dfrac{c}{a} = \dfrac{-24}{24} = -1$

Lines with $m _1$ and $m _2$ are slope are perpendicular.

Tangents from origin are at right angles to each other.

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

If two tangents inclined at an angle $\displaystyle 60^{\circ}$ are drawn to a circle of radius 3 cm then length of each tangent is equal to

  1. $\displaystyle \frac{3}{2}\sqrt{3}cm$
  2. $6 cm$
  3. $3 cm$
  4. $\displaystyle 3\sqrt{3}cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let PA and PB are the tangents on the circle. $\angle APB = 60$. the radius of the circle with center at O be 3 cm.
The two tangents drawn to a circle from an external point are equally inclined to the segment joining the center to the point.
Thus, $\angle APO = 30^{\circ}$
In $\triangle OAP$
$\angle OAP = 90^{\circ}$       ...(Angle between tangent and radius)
$\tan 30 = \cfrac{1}{\sqrt{3}} = \dfrac{OA}{AP}$
$PA = 3 \sqrt{3}$

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

If $5x-12y+10=0$ and $12y-5x+16=0$ are two tangents
to a circle then radius of the circle is

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

$5x-12y+10=0$ and $12y-5x+16=0$ are two parallel tangent to a circle.
Then distance $bet^{n}$ this two parallel tangents is $2r$.
$\therefore d=\left | \dfrac{-10-16}{\sqrt{5^{2}+12^{2}}} \right |=\left | \dfrac{26}{13} \right |=2$
$\therefore \ d=2r=2$
$\Rightarrow r=radius=1$

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

If ${ \theta } _{ 1 },{ \theta } _{ 2 }$ be the inclinations of tangents drawn from the point $P$ to the circle ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ and $\cot { { \theta  } _{ 1 } } +\cot { { \theta  } _{ 2 } } =k$, then the locus of $P$ is

  1. $k\left( { y }^{ 2 }+{ a }^{ 2 } \right) =2xy$
  2. $k\left( { y }^{ 2 }-{ a }^{ 2 } \right) =2xy$
  3. $k\left( { y }^{ 2 }+{ a }^{ 2 } \right) =4xy$

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

Equation of the circle is ${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$    ...(1)

Let $P$ be the point $\left( { x } _{ 1 },{ y } _{ 1 } \right) $.
Equation of any tangent to (1) is $y=mx+a\sqrt { 1+{ m }^{ 2 } } $
It is passes through $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $, then
${ y } _{ 1 }=m{ x } _{ 1 }+a\sqrt { 1+{ m }^{ 2 } } \Rightarrow { y } _{ 1 }-m{ x } _{ 1 }=a\sqrt { 1+{ m }^{ 2 } } $
Squaring ${ { y } _{ 1 } }^{ 2 }+2mx _{ 1 }{ y } _{ 1 }+{ m }^{ 2 }{ { x } _{ 1 } }^{ 2 }={ a }^{ 2 }\left( 1+{ m }^{ 2 } \right)$
$ \Rightarrow \left( { { x } _{ 1 } }^{ 2 }-{ a }^{ 2 } \right) { m }^{ 2 }-2{ x } _{ 1 }{ y } _{ 1 }m+\left( { { y } _{ 1 } }^{ 2 }-{ a }^{ 2 } \right) =0$   ...(2)
This is a quadratic in $m$. If ${ m } _{ 1 }$ and ${ m } _{ 2 }$ are its roots, then these are the slopes of the tangents from $P$.
Since inclination of tangents are given to be ${\theta} _{1}$ and ${\theta} _{2}$
$\therefore$ Let ${ m } _{ 1 }=\tan{{\theta} _{1}}$ and ${ m } _{ 2 }=\tan{{\theta} _{2}}$ 
$\displaystyle \Rightarrow \frac { 1 }{ { m } _{ 1 } } +\frac { 1 }{ { m } _{ 2 } } =k\Rightarrow { m } _{ 1 }+{ m } _{ 2 }=k{ m } _{ 1 }{ m } _{ 2 }$
$\displaystyle \therefore \frac { 2{ x } _{ 1 }{ y } _{ 1 } }{ { { x } _{ 1 } }^{ 2 }-{ a }^{ 2 } } =k.\frac { { { y } _{ 1 } }^{ 2 }-{ a }^{ 2 } }{ { { x } _{ 1 } }^{ 2 }-{ a }^{ 2 } } \Rightarrow 2{ x } _{ 1 }{ y } _{ 1 }=k\left( { { y } _{ 1 } }^{ 2 }-{ a }^{ 2 } \right) $
$\therefore $ Locus of $P$ is $k\left( { y }^{ 2 }{ -a }^{ 2 } \right) =2xy$

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 tangents from the origin to the circle $(x-7)^{2}+(y+1)^{2}=25$ is

  1. $\displaystyle \frac{\pi}{3}$
  2. $\displaystyle \frac{\pi}{6}$
  3. $\displaystyle \frac{\pi}{2}$
  4. $\displaystyle \frac{\pi}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(x-7)^2+(y+1)^2=25$
PA=PB=length of tangent from $(0,0) \space  to \space  (x-7)^2+(y+1)^2-25=0$
$=\sqrt{51}$
$\Rightarrow PA=PB=\sqrt{7^2+1-25}=5$
In $\Delta  OAP,$
$\tan  \alpha =\dfrac{OA}{PA}=\dfrac{5}{5}=1$
$\alpha =45^{\circ}$
So, angle both tangents $ =2\alpha =90^{\circ}$

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

State true or false
The length of tangent from an external point on a circle is always greater than the radius of the circle.

  1. True

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

false, it is not always required it can even be less or greater than the radius of the circle, it depend on how far the point is from the center of the circle. 

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

State true or false
The length of tangent from an external point P on a circle with centre O is always less than OP.

  1. True

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

true 

since if the tangent intersects the circle at Q the PQO forms a right angled triangle with hypotenuse PO so the length PQ is always less than PO as hypotenuse is the largest in a triangle.

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

Two tangents are drawn to a circle and the angle between them is $\displaystyle { 30 }^{ \circ  }$. What is the angle between the radii that are drawn at the point of contact of these two tangents.

  1. $\displaystyle { 30 }^{ \circ }$
  2. $\displaystyle { 60 }^{ \circ }$
  3. $\displaystyle { 90 }^{ \circ }$
  4. $\displaystyle { 150 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The angle between the two tangents and the angle between the radii at the points of contact are supplementary, as they form a quadrilateral with two 90-degree angles. Thus, 180 - 30 = 150 degrees.