Quantitative Aptitude

Geometry

1,950 Questions
All Circle Tangents and Chords 772 Coordinate Geometry 360 Circle and Arc Measures 216 Triangle Centers and Medians 175 Composite and Inscribed Shapes 174 Similarity, Congruence, and Proportionality 118 Right Triangles and Pythagorean Theorem 100 Standard 2D Mensuration 22 Polygon and Angle Sums 10 3D Solid Mensuration 3

Geometry Questions

Multiple choice

Consider the circles $C_1\equiv {x}^2 + y^2- 4 = 0 $, $C_2\equiv {x}^2 + y^2- 6x +8= 0 $,$C_3\equiv {x}^2 + y^2- 8x- 2y + 16 = 0 $. The number of common tangents that can be drawn to touch at least two of the circle is

  1. 8

  2. 9

  3. 10

  4. 12
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circles are C1: x^2 + y^2 = 4 (center (0,0), r=2), C2: (x-3)^2 + y^2 = 1 (center (3,0), r=1), and C3: (x-4)^2 + (y-1)^2 = 1 (center (4,1), r=1). By calculating the distances between centers and comparing them to the sum/difference of radii, one can determine the number of common tangents for each pair. C1 and C2 are externally tangent (3 common tangents), C2 and C3 are separate (4 common tangents), and C1 and C3 are separate (4 common tangents). The total number of unique common tangents is 8.

Multiple choice

Write True or False and justify your answer in each of the following : If angle between two tangents drawn from a point P to a circle of radius a and centre O is $90^{0}$, then OP$= a\sqrt{2}$.

  1. True

  2. False

  3. Ambiguous

  4. Data insufficient
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The tangents from an external point P to a circle are equal in length. The radius to the point of tangency is perpendicular to the tangent. This forms a square with the radius and the tangents, where the distance OP is the diagonal of a square with side a, so OP = a*sqrt(2).

Multiple choice

$OA.OB$ are the radii of a circle with $O$ as centre, the angle $AOB={120}^{o}$. Tangents at $A$ and $B$ are drawn to meet in the point $C$. If $OC$ intersects the circle in the point $D$, then $D$ divides $OC$ in the ratio.
  1. $1:2$
  2. $1:3$
  3. $1:1$
  4. $2:3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Because the tangents at A and B are equal, OC bisects angle AOB and is perpendicular to the chord AB. With angle AOB = 120 degrees, the distance from O to the tangent intersection C is twice the radius. The circle point D lies one radius from O, so OD equals DC and the ratio is 1:1.

Multiple choice

A tangent is drawn to a circle from a point $P$ which is at a distance of $13$ cm from the center of the circle of radius $5 cm.$ The length of the tangent is?
  1. $10cm$
  2. $12cm$
  3. $18cm$

  4. None of the options
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The radius, the tangent, and the line from the center to the external point form a right-angled triangle. By the Pythagorean theorem, the tangent length is sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12 cm.

Multiple choice

Two circles with radii '$a$' and '$b$' respectively touch each other externally. Let '$c$' be the radius of a circle that touches these two circles as well as a common tangent to the two circles. Then
  1. $-\dfrac {1}{\sqrt a}-\dfrac {1}{\sqrt b}=\dfrac {1}{\sqrt c}$
  2. $\dfrac {1}{\sqrt a}-\dfrac {1}{\sqrt b}=-\dfrac {1}{\sqrt c}$
  3. $\dfrac {1}{\sqrt a}+\dfrac {1}{\sqrt b}=\dfrac {1}{\sqrt c}$
  4. $\dfrac {1}{\sqrt a}-\dfrac {1}{\sqrt b}=\dfrac {1}{\sqrt c}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For two externally touching circles of radii a and b, a third circle of radius c touching both and their common tangent satisfies the Descartes circle theorem variant for a line (which acts as a circle of infinite radius). The relationship is 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b).

Multiple choice

$X$ is a point outside of a circle whose center is $O$. From $X$, a tangent whose length is $a$ is drawn to the circle and the shortest distance from $X$ to the circle is $\dfrac{a}{2}$. Find the radius of the circle.
  1. $\dfrac{3a}{4}$
  2. $\dfrac{3a}{2}$
  3. $\dfrac{1}{2}a$
  4. $a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let R be the radius. Distance from center O to X is d = sqrt(a^2 + R^2). Shortest distance to circle is d - R = a/2. So d = R + a/2. Substituting: (R + a/2)^2 = a^2 + R^2. R^2 + aR + a^2/4 = a^2 + R^2. aR = 3a^2/4. R = 3a/4.

Multiple choice

Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius $r$. If PS and RQ intersect at a point X on the circumference of the circle, then $2r$ equals
  1. $\sqrt{PQ.RS}$
  2. $\dfrac{PQ+RS}{2}$
  3. $\dfrac{2PQ.RS}{PQ+RS}$
  4. $\sqrt{\dfrac{PQ^2+RS^2}{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In this configuration, the tangents PQ and RS at the ends of diameter PR create properties where the triangles formed are similar. Specifically, the geometry leads to the relation 2r = sqrt(PQ * RS).

Multiple choice

$BC, AB$ and $AC$ are tangents to the circle at $D, E$ and $F$ respectively. $\angle EBD=x^{\circ}, \angle FCD=y^{\circ}.$ Then $\angle EDF$ equals
  1. $x+y$
  2. $\displaystyle \frac{x}{2}-y$
  3. $90^{\circ}-\left ( x+y \right )$
  4. $\displaystyle \frac{x+y}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the properties of tangents from an external point, the angles in the triangle formed by the circle and the tangents relate to the center. The angle EDF is half the sum of the angles at the vertices B and C in the triangle ABC.