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

The lengths of tangents from two given points to a given circle are $\displaystyle t_{1}: &: t_{2}$. If the two points are conjugate to each other w.r.to the given circle the distance between the points will be
  1. $\displaystyle \sqrt{\dfrac{t_{1}^{2}+t_{2}^{2}}{2}}$
  2. $\displaystyle \sqrt{t_{1}^{2}+t_{2}^{2}}$
  3. $\displaystyle \sqrt{t_{1}t_{2}}$
  4. $\displaystyle \dfrac{2t_{1}t_{2}}{t_{1}+t_{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For conjugate points with respect to a circle, the distance d between them satisfies d^2 = t1^2 + t2^2, where t1 and t2 are the lengths of the tangents from the points to the circle.

Multiple choice

Let C be a circle centre O. Let T be a point on the circle and P a point outside the circle such that PT is tangent to C. Assume that the segment OP intersects C in a point Q If PT = 12 and PQ = 8 the radius of C is
  1. $r = 40$
  2. $r = 5$
  3. $\displaystyle r=4\sqrt{5}$
  4. $\displaystyle r=4\sqrt{13}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By the power of a point theorem, PT^2 = PQ * (PQ + 2r). Given PT = 12 and PQ = 8, 144 = 8 * (8 + 2r). 18 = 8 + 2r, so 10 = 2r, r = 5.

Multiple choice

lf the area of the quadrilateral formed by the tangent from the origin to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}+6\mathrm{x}- 1 \mathrm{0}\mathrm{y}+\mathrm{c}=0$ and the pair of radii at the points of contact of these tangents to the circle is $8$ square units, then $c$ is a root of the equation
  1. $\mathrm{c}^{2}-32\mathrm{c}+64=0$
  2. $\mathrm{c}^{2}-34\mathrm{c}+64=0$
  3. $\mathrm{c}^{2}+2\mathrm{c}-64=0$
  4. $\mathrm{c}^{2}+34\mathrm{c}-64=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a circle x^2 + y^2 + 2gx + 2fy + c = 0, the radius r = sqrt(g^2 + f^2 - c). Here g=3, f=-5, so r = sqrt(9 + 25 - c) = sqrt(34 - c). The distance from origin to center is d = sqrt(3^2 + (-5)^2) = sqrt(34). The area of the quadrilateral formed by tangents and radii is r * sqrt(d^2 - r^2). Given area = 8, so r * sqrt(34 - r^2) = 8. Squaring gives r^2 * (34 - r^2) = 64. Substituting r^2 = 34 - c, we get (34 - c) * (34 - (34 - c)) = 64, which simplifies to (34 - c) * c = 64, or c^2 - 34c + 64 = 0.

Multiple choice

The lengths of the tangents from $(1,0), (2,0), (3,2)$ to a circle are $1, \sqrt{7},\sqrt{2}$. Then the equation of the circle is
  1. $\mathrm{x}^{2}+\mathrm{y}^{2}+6\mathrm{x}+17\mathrm{y}+6=0$
  2. $2\mathrm{x}^{2}+2\mathrm{y}^{2}+6\mathrm{x}-17\mathrm{y}-6=0$
  3. $2\mathrm{x}^{2}+2\mathrm{y}^{2}-6\mathrm{x}-17\mathrm{y}-6=0$
  4. $2\mathrm{x}^{2}+2\mathrm{y}^{2}-6\mathrm{x}-17\mathrm{y}+6=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The length of the tangent from (x1, y1) to circle x^2 + y^2 + 2gx + 2fy + c = 0 is sqrt(x1^2 + y1^2 + 2gx1 + 2fy1 + c). Using the given points, we set up a system of equations to solve for g, f, and c.

Multiple choice

The equation of pair of tangents from origin to a circle is $24xy + 7y^{2} = 0$. If the radius of the circle is 3 , then the length of the tangent drawn from the origin is
  1. $7$
  2. $24$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of the pair of tangents is 24xy + 7y^2 = 0, which factors to y(24x + 7y) = 0. This gives two lines: y = 0 and y = -24/7 x. The distance from the center (h, k) to these lines must equal the radius 3. Solving for the center and then the distance from the origin yields 4.

Multiple choice

Consider a family of circles which are passing through $M(1,1)$ and are tangent to x-axis. If $(h,k)$ is the centre of circle, then
  1. $\quad k\ge \cfrac { 1 }{ 2 } $
  2. $-\cfrac { 1 }{ 2 } \le k\le \cfrac { 1 }{ 2 } $
  3. $k\le \cfrac { 1 }{ 2 } $
  4. $0 < k < \cfrac { 1 }{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle passes through (1,1) and is tangent to the x-axis, so its radius is |k|. The equation is (x-h)^2 + (y-k)^2 = k^2. Substituting (1,1): (1-h)^2 + (1-k)^2 = k^2, which simplifies to (1-h)^2 + 1 - 2k + k^2 = k^2, so (1-h)^2 + 1 = 2k. Since (1-h)^2 >= 0, 2k >= 1, so k >= 1/2.

Multiple choice

If distance between centres of two circles is $a$ and radii are $r_{1}$and $r_{2}$ , then length of common tangent is
  1. $r_{1}+r_{2}$
  2. $a^{2}$
  3. $\sqrt{a^{2}-(r_{1}+r_{2})^{2}}$

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

The length of the transverse (or indirect) common tangent between two circles with radii r1 and r2 and distance between centers a is given by the formula sqrt(a^2 - (r1 + r2)^2). This is derived using the Pythagorean theorem on the right triangle formed by the centers and the parallel shift of the tangent.