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

$\overline { M N }$ and $\overline { M Q }$ are two tangents from a point $M$ to a circle with centre $0$ If $m \angle N O Q = 120 ^ { \circ } ,$ then ?

  1. $N Q = M N = M Q$
  2. $N Q = O M$
  3. $O Q = O M$
  4. $O N = M N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If angle NOQ = 120 degrees, then in the quadrilateral MONQ, the angles at N and Q are 90 degrees. Thus, angle M = 180 - 120 = 60 degrees. Triangle MNQ is isosceles with angle M = 60, so it is equilateral, meaning NQ = MN = MQ.

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

The chord of contact of the pair of tangents to the circle $x^2+y^2=1$ drawn from any point on the line $2x+y=4$ passes through a fixed point. 

  1. True

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

If chords are drawn to the circle from a fixed point $(x _1,y _1)$ and then tangents are drawn at point of contact, the point of intersection of all tangents lie on a fixed point.


The fixed point is called pole and fixed line is called polar.


Equation of polar is $T=0$.

$C:x^2+y^2-1=0$

Equation of polar is $T=0$.

$xx _1+yy _1-1=0$

The line is identical to given line $2x+y-4=0$.

By comparing coefficients, we get,
$\dfrac{x _1}{2}=\dfrac{y _1}{1}=\dfrac{-1}{-4}$

$x _1=\dfrac{1}{2},y _1=\dfrac{1}{4}$

Hence, the fixed point is $(\dfrac{1}{2}, \dfrac{1}{4})$.

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

From a point $P$ which is at a distance of $13$ cm from the centre $O$ of a circle of radius $5$ cm, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is:

  1. $60$ cm$^{2}$
  2. $65$ cm$^{2}$
  3. $30$ cm$^{2}$
  4. $32.5$ cm$^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The radius perpendicular tangent at the pt. of contact, therefore, $OQ\perp PQ$ and $OR\perp PR$
In rt. $\triangle OPQ$, we have
$PQ=\sqrt{OP^{2}-OQ^{2}}$
   $=\sqrt{169-25}=\sqrt{144}=12$ cm
$\Rightarrow $ $PR=12$ cm (Two tangents from the same external pt. to a circle are equal)
Now area of quad. $PQOR=2\times $Area of $\triangle POQ$
   $\displaystyle =\left ( 2\times \frac{1}{2}\times 12\times 5 \right )$ cm$^{2}=60$ cm$^{2}$

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

Circles ${ C } _{ 1 },{ C } _{ 2 },{ C } _{ 3 }$ have their centres at $\left( 0,0 \right) ,\left( 12,0 \right) ,\left( 24,0 \right) $ and have radii $1,2$ and $4$ respectively. Line ${t} _{1}$ is a common internal tangent to ${C} _{1}$ and ${C} _{2}$ and has a positive slope and line ${t} _{2}$ is a common internal tangent to ${C} _{2}$ and ${C} _{3}$ and has a negative slope. Given that lines ${t} _{1}$ and ${t} _{2}$ intersect at $(x,y)$ and that $x=p-q\surd r$, where $p,q$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r$.

  1. $p+q+r=26$
  2. $p+q+r=24$
  3. $p+q+r=28$
  4. $p+q+r=27$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

For the two circles ${ x }^{ 2 }+{ y }^{ 2 }=16$ and ${ x }^{ 2 }+{ y }^{ 2 }-2y=0$ there is/are

  1. One pair of common tangents

  2. Only one common tangent

  3. Three common tangents

  4. No common tangent
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The centres and radii of given circles are ${ C } _{ 1 }\left( 0,0 \right) ,{ r } _{ 1 }=4$ and ${ C } _{ 2 }\left( 0,1 \right) ,{ r } _{ 2 }=\sqrt { 0+1 } =1$
Now, ${ C } _{ 1 }{ C } _{ 2 }=\sqrt { 0+{ \left( 0-1 \right)  }^{ 2 } } =1$
and ${ r } _{ 1 }-{ r } _{ 2 }=4-1=3$
$\therefore { C } _{ 1 }{ C } _{ 2 }<{ r } _{ 1 }-{ r } _{ 2 }$
Hence, second circle lies inside the first circle.

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

From a point outside a circle, one tangent and one secant are drawn. The length of exterior part of secant is $7$ cm and that of interior part is $9$ cm. Find the length of tangent segment.

  1. $10.6$ cm
  2. $10.9$ cm
  3. $11.2$ cm
  4. $11.6$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let length of tangent be $l$

Length of exterior part of secant $=m=7 $ cm
Length of interior part of secant $=n=9 $ cm
Now using the secant intersection theorem, we have
${ l }^{ 2 }=m(m+n)\ \Rightarrow { l }^{ 2 }=7(7+9)\ \Rightarrow { l }^{ 2 }=112 $
$\Rightarrow l=\sqrt { 112 } =10.6$ cm
Option A is correct.

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

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is $60^0$. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

  1. 4 cm

  2. 6 cm

  3. 8 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Join O and P.
Now, in triangles OPQ and OPR,
OP = OP (Common)
OQ = OR (radius of circle)
PQ = PR (tangents from single point)

Hence OPQ and OPR are congruent triangles.
$\angle OPQ = \angle OPR = 30^{\circ}$


Thus in triangle OPQ, $\dfrac{OQ}{OP} = Sin 30$

OP = $\dfrac{4}{Sin30}$

OP = 8 cm

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

If the angle between two radii of a circle is $140^{\circ}$, then the angle between the tangents at the ends of the radii is :

  1. $90^{\circ}$
  2. $40^{\circ}$
  3. $70^{\circ}$
  4. $60^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since tangents and radii are perpendicular at the point of contact, in the quadrilateral formed by the two radii and the tangents at their ends, we have two right angles at the two points of contacts.

Let the angle between the tangents be $x^o$. Then
$140 + 90 + 90 + x = 360 \Rightarrow x = 40^o$.
So option B is the right answer.

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 of $60^{\circ}$ are drawn to a circle of radius 3 cm, then the length of each tangent is equal to:

  1. $\dfrac{3\sqrt{3}}{2}$ cm
  2. $2\sqrt{3}$ cm
  3. $3\sqrt{3}$ cm

  4. 6 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Tangent is perpendicular to radius at the point of contact.

By symmetry with respect to the line joining the center and the point from which tangents are drawn, we have the length of tangent $=\dfrac{3}{\tan 30^o}=3\sqrt{3} cm$.
So option C is the right answer.

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

From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn. Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n$. It is found that $t$, the length of the tangent, is the mean proportional between $m$ and $n$. If $m$ and $t$ are integers, then $t$ may have the following number of values.

  1. Zero

  2. One

  3. Two

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

Circumference = 10 units

m+n=10
n=10-m
't' is the length of the tangent.
$t^{2}=mn$
$t=\sqrt{m(10-m)}$
At $m=1, t=3$
At $m=2, t=4$
$\therefore$ Two values are possible for t.

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

A pair of tangents are drawn from a point $P$ to the circle $x^{2} + y^{2} = 1$. If the tangents make an intercept of $2$ on the line $x = 2$, the locus of $P$ is

  1. Straight line

  2. Pair of lines

  3. Circle

  4. Parabola
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $P$ be $(h, k)$ then by $SS _{1} = T^{2}$ the equation of pair of tangents drawn from $P$ to $x^{2} + y^{2} = 1$ is
$(h^{2} + k^{2} - 1)(x^{2} + y^{2} - 1) = (hx + ky - 1)^{2}$
Its intersection with the line $x = 1$ is given by
$y^{2}(h^{2} + k^{2} - 1) = (ky + h - 1)^{2}$
or $y^{2}(h^{2} - 1) - 2yk (h - 1) - (h - 1)^{2} = 0 .....(1)$
It is a quadratic in $y$ and we are given that length of intercept is $1 \therefore y _{1} - y _{2} = 2$
or $(y _{1} + y _{2})^{2} - 4y _{1}y _{2} = 1$
or $\left [\dfrac {2k(h - 1)}{h^{2} - 1}\right ]^{2} + 4\dfrac {(h - 1)^{2}}{h^{2} - 1} = 4$
or $\dfrac {4k^{2}}{(h + 1)^{2}} + \dfrac {4(h - 1)}{h + 1} = 4$
or $k^{2} = (h + 1)^{2} - (h^{2} - 1) = 2h + 2$
Hence the locus of $(h, k)$ is $y^{2} =2(x + 1)$ which represents a parabola.

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

A tangent from $P$, a point in the exterior of a circle touches circle at $Q$. If $OP=13$, $PQ=5$, then the diameter of the circle is ______________

  1. $576$
  2. $15$
  3. $8$
  4. $24$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since tangent is perpendicular to the radius through the point of contact
so, $PQ \bot OQ$

therefore
${\left( {PQ} \right)^2} + {\left( {OQ} \right)^2} = {\left( {OP} \right)^2}$

$ = {\left( 5 \right)^2} + {r^2} = {\left( {13} \right)^2}$

$ = {r^2} = 169 - 25$

$\Rightarrow {r^2} = 144$

$\Rightarrow r = 12cm$

so, diameter of the circle $2 \times r$
$=2 \times 12$ $=24cm$