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

Let a circle be given by $2x(x-a)+y(2y-b)=0;$ $(a\ne 0, b\ne 0)$. Find the condition on $a$ and $b$ if two chords, each bisected by the $x$-axis, can be drawn to the circle from the point $\left(a,\cfrac{b}{2}\right)$.
  1. ${a}^{2}>2{b}^{2}$
  2. $2{a}^{2}>{b}^{2}$
  3. ${a}^{2}<2{b}^{2}$
  4. $2{a}^{2}<{b}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle equation is 2x^2 - 2ax + 2y^2 - by = 0. The center is (a/2, b/4). A chord bisected by the x-axis means the midpoint of the chord has y=0. For two such chords to exist from (a, b/2), the point must lie outside the circle, and the geometry of the circle must allow for the intersection of the chord with the x-axis.

Multiple choice

State the following statement is true or false The measure of the line segment joining the centre of a circle to the mid-point of a chord is $\dfrac{\sqrt{4r^{2}-L^{2}}}{2}$

  1. True

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

The line segment from the center of a circle to the midpoint of a chord is perpendicular to the chord, forming a right triangle. The hypotenuse of this triangle is the radius r, and one leg is half the length of the chord, L/2. Applying the Pythagorean theorem, the distance d satisfies d^2 + (L/2)^2 = r^2, which simplifies to d = sqrt(4r^2 - L^2) / 2.

Multiple choice

A circle is inscribed into a rhombus ABCD with one angle $\displaystyle 60^{\circ} .$ The distance from the centre of the circle to the nearest vertex is equal to $1$. If P is any point of the circle, then $\displaystyle \left | PA \right |^{2}+\left | PB \right |^{2}+\left | PC \right |^{2}+\left | PD \right |^{2}$ is equal to
  1. $12$
  2. $11$
  3. $9$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a rhombus with a 60-degree angle, the distance from the center to the vertex is related to the geometry of the inscribed circle. Using coordinates or geometric properties, the sum of squared distances from any point on the circle to the vertices of the rhombus is constant and equals 11.

Multiple choice

If $ax + by - 3 = 0$ is the equation of the shortest chord of the circle $\displaystyle \left ( x-3 \right )^{2}+\left ( y-4 \right )^{2}=4$ passing through the point $(2,3)$, then $|a + b|$ is
  1. $\dfrac{6}{5}$
  2. $4$
  3. $1$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shortest chord passing through (2,3) is the one perpendicular to the radius at (2,3). Center is (3,4). Radius vector from (3,4) to (2,3) has slope (3-4)/(2-3) = -1/-1 = 1. The chord must have slope -1. Equation: y - 3 = -1(x - 2) => y - 3 = -x + 2 => x + y - 5 = 0. Comparing to ax + by - 3 = 0, we need to scale to match constant term. Multiply by 3/5: (3/5)x + (3/5)y - 3 = 0. So a=3/5, b=3/5. |a+b| = 6/5.

Multiple choice

Length of the common chord of the circles $\displaystyle \left ( x-1 \right )^{2}+\left ( y+1 \right )^{2}=c^{2}: : and: : \left ( x+1 \right )^{2}+\left ( y-1 \right )^{2}=c^{2}$ is
  1. $\displaystyle \frac{1}{2}\sqrt{c^{2}-2}$
  2. $\displaystyle \sqrt{c^{2}-2}$
  3. $\displaystyle 2\sqrt{c^{2}-2}$
  4. $(c + 2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The centers of the circles are (1, -1) and (-1, 1). The distance between centers is sqrt(2^2 + (-2)^2) = sqrt(8) = 2*sqrt(2). The common chord length formula for two circles of radius c is 2*sqrt(c^2 - d^2/4), where d is the distance between centers. Substituting d = 2*sqrt(2), we get 2*sqrt(c^2 - 8/4) = 2*sqrt(c^2 - 2).

Multiple choice

The lengths of two parallel chords of a circle $1cm$ apart are respectively $10cm$ and $8cm$. The distance of the longer chord from the centre of the circle is-
  1. $4cm$
  2. $5cm$
  3. $7cm$
  4. $12cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the radius be r. Distance of 10cm chord from center is sqrt(r^2 - 5^2). Distance of 8cm chord is sqrt(r^2 - 4^2). The difference is 1cm. sqrt(r^2 - 16) - sqrt(r^2 - 25) = 1. Solving gives r^2 = 41. Distance of 10cm chord = sqrt(41 - 25) = sqrt(16) = 4cm.

Multiple choice

If two chords of lengths $2a$ each, of a circle of radius $R$, intersect each other at right angles then the distance of their point of intersection from the centre of the circle is
  1. $2\sqrt{R^2-a^2}$
  2. $\sqrt{2(R^2-a^2)}$
  3. $4\sqrt{(R^2-a^2)}$
  4. $2(R^2-a^2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a chord of length 2a, the distance from the center is sqrt(R^2 - a^2). If two such chords intersect at right angles, they form a square with the center, where the distance from the center to the intersection point is the diagonal of a rectangle with sides equal to the distances of the chords from the center. Thus, distance = sqrt((sqrt(R^2-a^2))^2 + (sqrt(R^2-a^2))^2) = sqrt(2(R^2-a^2)).