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 general knowledge

find the ODD one.

  1. diameter

  2. diagonal

  3. tangent

  4. arc
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Option B is the odd one out. Diameter, tangent, and arc are all terms related to circles, while diagonal is a term related to polygons and other shapes connecting non-adjacent vertices. A circle does not have a diagonal.

Multiple choice general knowledge math & puzzles

A line segment that is tangent to a circle O at its midpoint has its endpoints at A and B. If AB = the diameter of circle O, then triangle OAB is a right triangle.

  1. True

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

The radius drawn to a point of tangency is perpendicular to the tangent line. Since the tangent is perpendicular to radius at the midpoint, and AB is a line segment, angle OAB is 90°, making triangle OAB a right triangle.

Multiple choice

On a map drawn on Gnomonic projection, the great circle course between two points is shown by

  1. a sine curve

  2. a straight line

  3. a parabola

  4. an arc of a circle
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Gnomonic projection is unique because all great circles appear as straight lines on this projection. This property makes it useful for navigation and plotting the shortest distance routes between two points on Earth's surface, since great circles represent the shortest path on a sphere.

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

Tangents $PA$ and $PB$ drawn to ${ x }^{ 2 }+{ y }^{ 2 }=9$ from any arbitrary point $'P'$ on the line ${ x }+{ y }=25$. Locus of midpoint of chord $AB$ is

  1. $25({ x }^{ 2 }+{ y }^{ 2 })=9(x+y)$
  2. $25({ x }^{ 2 }+{ y }^{ 2 })=3(x+y)$
  3. $5({ x }^{ 2 }+{ y }^{ 2 })=3(x+y)$

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

Let the point on the line $x+y=25$ be $P(a,b)$
Thus equation of chord of contact $AB$ from point $P$ to the circle is given by,
$T  =0 \Rightarrow ax+by = 9$  (i)
Let mid point of $AB$ be $R(h,k)$.
Now equation of chord $AB$ with mid point $R$ is given by,
$T = S _1 \Rightarrow hx+ky = h^2+k^2$ (ii)
Both line (i) and (ii) represents the same line $AB$
$\therefore \displaystyle \frac{a}{h}=\frac{b}{k} = \frac{9}{h^2+k^2}$
$\Rightarrow  a=\cfrac{9h}{h^2+k^2}, b = \cfrac{9k}{h^2+k^2}$
Also point $(a,b)$ lie on the line $x+y = 25$
$\Rightarrow a+b = 25\Rightarrow \cfrac{9h}{h^2+k^2}+ \cfrac{9k}{h^2+k^2}=25$
$ \Rightarrow 25(h^2+k^2) = 9(h+k)$
Hence required locus of $R(h,k)$ is given by $25(x^2+y^2) = 9(x+y)$

Multiple choice maths geometric constructions circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

Given are the steps are construction of a pair of tangents to a circle of radius $4$cm from a point on the concentric circle of radius $6$cm. Find which of the following step is wrong?
(P) Take a point O on the plane paper and draw a circle of radius OA$=4$cm. Also, draw a concentric circle of radius OB$=6$cm.
(Q) Find the mid-point A of OB and draw a circle of radius BA$=$AO. Suppose this circle intersects the circle of radius $4$cm at P and Q.
(R) Join BP and BQ to get the desired tangents from a point B on the circle of radius $6$ cm.

  1. Only (P)

  2. Only (Q)

  3. Both (P) & (Q)

  4. Both (Q) & (R)
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths geometric constructions circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

What are the tools required for constructing a tangent to a circle?

  1. ruler

  2. compass

  3. pencil

  4. all the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tools required for constructing a tangent to a circle is ruler, compass and pencil.

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

Tangents PA and PB drawn to $ x^2+y^2=9 $ from any arbitrary point 'P ' on the line $ x+y=25 $. Locus of midpoint of chord AB is

  1. $ 25(x^2+y^2)=9(x+y) $
  2. $ 25(x^2+y^2)=3(x+y) $
  3. $ 5(x^2+y^2)=3(x+y) $

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

Let the point on the line $x+y=25$ be $P(a,b)$
Thus equation of chord of contact AB from point P to the circle is given by,
$T  =0 \Rightarrow ax+by = 9$  (i)
Let mid point of AB be $R(h,k)$.
Now equation of chord AB with mid point R is given by,
$T = S _1 \Rightarrow hx+ky = h^2+k^2$ (ii)
Both line (i) and (ii) represents the same line AB
$\therefore \displaystyle \frac{a}{h}=\frac{b}{k} = \frac{9}{h^2+k^2}$
$\Rightarrow  a=\cfrac{9h}{h^2+k^2}, b = \cfrac{9k}{h^2+k^2}$
Also point $(a,b)$ lie on the line $x+y = 25$
$\Rightarrow a+b = 25 \Rightarrow 25(h^2+k^2) = 9(h+k)$
Hence required locus of $R(h,k)$ is given by, $25(x^2+y^2) = 9(x+y)$

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.