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

Circles are drawn on chords of the rectangular hyperbola $xy=4$ parallel to the line $y=x$ as diameters.All such circles pass through two fixed points whose coordinates are 

  1. $\left(2,2\right)$
  2. $\left(2,-2\right)$
  3. $\left(-2,2\right)$
  4. $\left(-2,-2\right)$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation
Given:Rectangular hyperbola $xy=4={c}^{2}$
$\Rightarrow\,{c}^{2}=4$
$\Rightarrow\,c=2$
Let $P$ and $Q$ be the end points on the Rectangular hyperbola where $P\left(2{t} _{1},\dfrac{2}{{t} _{1}}\right)$ and $Q\left(2{t} _{2},\dfrac{2}{{t} _{2}}\right)$
Using the end points of diameter the equation of the circle

$C:\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$         ..........$(1)$

Now,Slope of the $PQ=\dfrac{\dfrac{2}{{t} _{2}}-\dfrac{2}{{t} _{1}}}{2{t} _{2}-2{t} _{1}}$

$=\dfrac{2\left(\dfrac{1}{{t} _{2}}-\dfrac{1}{{t} _{1}}\right)}{2\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{-1}{{t} _{1}{t} _{2}}$

Hence $PQ$ is the diameter for circle and it is parallel to the line $y=x$

Slope of $PQ=$Slope of the line $y=x$

$\Rightarrow\,\dfrac{-1}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{t} _{1}{t} _{2}=-1$

$(1)\Rightarrow\,\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$ 

$\Rightarrow\,x\left(x-2{t} _{2}\right)-2{t} _{1}\left(x-2{t} _{2}\right)+y\left(y-\dfrac{2}{{t} _{2}}\right)-\dfrac{2}{{t} _{1}}\left(y-\dfrac{2}{{t} _{2}}\right)=1$
 
$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4{t} _{1}{t} _{2}+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4\times -1+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{\times -1}=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-2x\left({t} _{2}+{t} _{1}\right)-4-2y\left(\dfrac{1}{{t} _{2}}+\dfrac{1}{{t} _{1}}\right)-4=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{{t} _{1}{t} _{2}}\right)=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{-1}\right)=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)+2y\left({t} _{1}+{t} _{2}\right)=1$ 

$\Rightarrow\,{x}^{2}+{y}^{2}-8+\left(2y-2x\right)\left({t} _{2}+{t} _{1}\right)=1$ is of the form $C+\lambda\,L$ 

where $C={x}^{2}+{y}^{2}-8=0$ is the equation of a circle.
and $L=2y-2x=0$ is the equation of a line.
$\Rightarrow\,y-x=0$ or $x=y$

Substituting $x=y$ in the equation ${x}^{2}+{y}^{2}-8=0$ we get
$\Rightarrow\,2{x}^{2}-8=0$

$\Rightarrow\,2\left({x}^{2}-4\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+2\right)=0$

$\therefore\,x=2,-2$

$\Rightarrow\,y=2,-2$ since $x=y$

Hence the coordinates of the fixed points are $\left(2,2\right)$ and $\left(-2,-2\right)$ 
Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The intersection point of,a perpendicular on tangent of a hyperbola from the focus  and a tangent lies on 

  1. director circle

  2. auxillary circle

  3. nine point circle

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The intersection point of a perpendicular on tangent of a hyperbola from the focus and a tangent lies on: Auxiliary circle
Auxiliary circle of a hyperbola which is a circle described on the major axis of a hyperbola as its diameter.
Let the hyperbola be
$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
The equation of auxiliary circle is $x^2+y^2=a^2$
Take a point $P(x _1,y _1)$
Through $P$ draw a line perpendicular to major axis intersecting major axis in $N$ and auxiliary circle in $P'$.
The points $P$ and $P'$ are called as corresponding points on the hyperbola and auxiliary circle respectively.
This angle is known as the eccentric angle of the point $P$ on the hyperbola and auxiliary circle respectively.
Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

If pair of tangents are drawn from any point $(p)$ on the circle ${x^2} + {y^2} = 1$ to the hyperbola $\frac{{{x^2}}}{2} - \frac{{{y^2}}}{1} = 1$ such that locus of circumcenter of triangle formed by pair of tangents and chord of contact is ${\lambda _1}{x^2} - 2{\lambda _2}{y^2} = 2{\left( {\frac{{{x^2}}}{2} - {y^2}} \right)^2}$, then 

  1. ${\lambda _1} = 2,{\lambda _2} = 1$
  2. ${\lambda ^2} _1 + {\lambda ^2} _2 = 5$
  3. ${\lambda _1} = 1,{\lambda _2} = - 1$
  4. ${\lambda ^2} _1 + {\lambda ^2} _2 = 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard property of the locus of the circumcenter of the triangle formed by the pair of tangents and the chord of contact for a hyperbola.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The total number of real tangents that can be drawn to the ellipse $3x^{2}+5y^{2}=32$ and $25x^{2}+9y^{2}=450$ passing through $(3,5)$ is

  1. $0$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(3,5)$ lies on $25x^2+9y^2=450$

Therefore, one tangent can be drawn

and $(3,5)$ lies outside $3x^2+5y^2=32$ because $S _1>0$

Therefore, two tangents can be drawn.
So total 3 tangents

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

The range of values of $\lambda$ for which the circles $ { x }^{ 2 }+{ y }^{ 2 }=4$ and ${ x }^{ 2 }+{ y }^{ 2 }-2\lambda y+5=0$ have two common tangents only is-

  1. $\lambda \epsilon \left( -\sqrt { 5 } ,\sqrt { 5 } \right) $
  2. $\lambda <-\sqrt { 5 } or\quad \lambda >\sqrt { 5 }$
  3. $-\sqrt { 5 } <\lambda <1$

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

Two circles have two common tangents if the distance between their centers is less than the sum of their radii and greater than the difference of their radii. Here, centers are (0,0) and (0, lambda), radii are 2 and sqrt(lambda^2 - 5). The condition leads to the specified range.

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

Intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

  1. True

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

For any two parallel tangents to a circle, the segment of a third tangent intercepted between them subtends a 90-degree angle at the center because the radii to the points of tangency are perpendicular to the tangents.

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