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

Two circles, each radius $5$, have a common tangent at $(1,1)$ whose equation is $3x+4y-7=0$. Then their centre are
  1. $(4,-5),(-2,3)$
  2. $(4,-3),(-2,5)$
  3. $(4,5),(-2,-3)$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The centers lie on a line perpendicular to the tangent 3x + 4y - 7 = 0, which has slope -3/4. The normal line has slope 4/3 and passes through (1,1), so y - 1 = 4/3(x - 1) => 4x - 3y - 1 = 0. Checking the options, (4,5) and (-2,-3) satisfy this line equation and are distance 5 from (1,1).

Multiple choice

If a chord of a circle $x^{2}+y^{2}=32$ makes equal intercepts of length $l$ on the co-ordinate axes, then
  1. $\ell <8$
  2. $\ell <16$
  3. $\ell >8$
  4. $\ell >16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line making equal intercepts l on axes is x/l + y/l = 1, or x + y = l. The distance from the center (0,0) to this line must be less than the radius (sqrt(32)). Distance = |l| / sqrt(1^2 + 1^2) = l / sqrt(2). So l / sqrt(2) < sqrt(32) => l < sqrt(64) => l < 8.

Multiple choice

The locus of the mid-points of the chords of the circle $x^{2}+y^{2}-4x-2y-11=0$which subtend $60^{0}$ at the centre is
  1. $x^{2}+y^{2}-4x-2y-7=0$
  2. $x^{2}+y^{2}+4x+2y-7=0$
  3. $x^{2}+y^{2}-2x-4y-7=0$
  4. $x^{2}+y^{2}+2x+4y+7=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The chord subtends 60 degrees at the center, so the triangle formed by the center and the chord endpoints is equilateral. The distance from the center (2,1) to the midpoint of the chord is r * cos(30) = sqrt(16) * sqrt(3)/2 = 2 * sqrt(3). The locus of midpoints is a circle concentric with the original circle, with radius equal to this distance. Radius squared = 12. Equation: (x-2)^2 + (y-1)^2 = 12. x^2 - 4x + 4 + y^2 - 2y + 1 = 12. x^2 + y^2 - 4x - 2y - 7 = 0.

Multiple choice

Two parallel chords are drawn on the same side of the centre of a circle of radius R. It is found that they subtend and angle of $\theta $ and $2\theta $ at the centre of the circle. The perpendicular distance between the chords is
  1. $1R\sin \dfrac{{3\theta }}{2}\sin \dfrac{\theta }{2}$
  2. $\left( {1 + \cos \dfrac{\theta }{2}} \right)\left( {1 - 2\cot \dfrac{\theta }{1}} \right)R$
  3. $2R\sin \dfrac{{3\theta }}{4}\sin \dfrac{\theta }{4}$
  4. $2R\sin 3\theta $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The distance from the center to a chord subtending angle 2A is R*cos(A). For angles theta and 2*theta, the distances are R*cos(theta/2) and R*cos(theta). The difference is R(cos(theta/2) - cos(theta)). Using trigonometric identities, this simplifies to 2R*sin(3*theta/4)*sin(theta/4).

Multiple choice

If the chord of contact of the tangents from a point on the circle $x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$ to the circle $x ^ { 2 } + y ^ { 2 } = b$ touches the circle $x ^ { 2 } + y ^ { 2 } = c ^ { 2 } ,$ then $a , b ,$ care in
  1. $A G P$
  2. $A . P $
  3. $G . P $
  4. $H . P $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a circle centered at the origin, the chord of contact from a point at distance a to a circle of radius b lies at distance b^2/a from the origin. If this chord touches the circle of radius c, then b^2/a = c, so b^2 = ac. Therefore, a, b, c are in geometric progression.