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

A line meets the co-ordinate axes in A & B. A circle is circumscribed about the $\triangle \ OAB$ . If 6 & 3 are the distances of the tangent to the circle at the origin O from the points A and B respectively, the diameter of the circle is
  1. $\frac{15}{2}$

  2. 6

  3. 2

  4. 9
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the line be x/a + y/b = 1. The distance from origin to the line is h = 1/sqrt(1/a^2 + 1/b^2). The tangent distances are related to the intercepts. Using geometry of the circumcircle of the right triangle OAB, the diameter is the hypotenuse AB = sqrt(a^2 + b^2). Given the tangent distances 6 and 3, the diameter is 9.

Multiple choice

In $\Delta ABC, x,y$ and $z$ are the distances of incentre from angular points $A,B,$ and $C$ respectively. If $\displaystyle \frac{xyz}{abc}=\displaystyle \frac{\lambda r}{s}$, then $\lambda=$
  1. $1$
  2. $2$
  3. $3$

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

In a triangle, the distance from the incentre to the vertices are x = r/sin(A/2), y = r/sin(B/2), z = r/sin(C/2). The product xyz = r^3 / (sin(A/2)sin(B/2)sin(C/2)). Using the identity sin(A/2)sin(B/2)sin(C/2) = r/(4R), we get xyz = 4Rr^2. Also, abc = 4Rrs. Thus, xyz/abc = (4Rr^2) / (4Rrs) = r/s. Therefore, lambda = 1.

Multiple choice

The distance of the incentre of the triangle $ABC$ from the point $A$ is?
  1. $4 R \sin \dfrac{A}{2}$
  2. $4 R \sin \dfrac{B}{2} \sin \dfrac{C}{2}$
  3. $4R \cos \dfrac{A}{2}$
  4. $4 R \cos \dfrac{B}{2} \cos \dfrac{C}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The distance from a vertex A to the incentre I is given by r / sin(A/2), where r is the inradius. Using r = 4R sin(A/2) sin(B/2) sin(C/2), the distance AI = 4R sin(B/2) sin(C/2).

Multiple choice

The locus of the mid points of the chords of the circle $x^{2} + y^{2} + 4x - 6y - 12 = 0$ which subtend an angle of $\pi/3$ radians at its circumference is
  1. $(x + 2)^{2} + (y - 3)^{2} = 6.25$
  2. $(x - 2)^{2} + (y + 3)^{2} = 6.25$
  3. $(x + 2)^{2} + (y - 3)^{2} = 18.75$
  4. $(x + 2)^{2} + (y + 3)^{2} = 18.75$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle is (x+2)^2 + (y-3)^2 = 25. Radius R=5. A chord subtending pi/3 at circumference subtends 2*pi/3 at center. The distance d from center to chord is R*cos(pi/3) = 5 * 0.5 = 2.5. The locus of midpoints of chords at distance d from center is a circle with radius d, centered at the same point: (x+2)^2 + (y-3)^2 = (2.5)^2 = 6.25.

Multiple choice

The locus of the middle point of the chord of the circle ${x}^{2}+{y}^{2}-12x+4=0$, that subtends an angle $\frac{2\pi}{3}$ at the centre is
  1. ${x}^{2}+{y}^{2}-12x+4y+30=0$
  2. ${x}^{2}+{y}^{2}-12x+4y+31=0$
  3. ${x}^{2}+{y}^{2}+12x+4y+30=0$
  4. ${x}^{2}+{y}^{2}-12x+28=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The chord subtends 120 degrees at the center (6,0). The midpoint of the chord forms a right triangle with the center and the chord endpoint. The distance from center to midpoint is r * cos(60) = sqrt(32) * 1/2 = sqrt(8). The locus of the midpoint is a circle centered at (6,0) with radius sqrt(8). (x-6)^2 + y^2 = 8 -> x^2 - 12x + 36 + y^2 = 8 -> x^2 + y^2 - 12x + 28 = 0.

Multiple choice

The locus of the centre of the circles such that the point $(2, 3)$ is the midpoint of the chord $5x+2y=16$ is
  1. $2x-5y+11=0$
  2. $2x-5y-11=0$
  3. $2x+5y+11=0$
  4. $None$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The chord is 5x + 2y = 16. The center (h, k) of the circle must lie on a line perpendicular to the chord passing through the midpoint (2, 3). The slope of the chord is -5/2, so the perpendicular slope is 2/5. Equation: y - 3 = (2/5)(x - 2) => 5y - 15 = 2x - 4 => 2x - 5y + 11 = 0.

Multiple choice

A triangle has a fixed base AB that is 2 cm long. The median from A to side BC is $1\frac{1}{2}$ cm long and can have any position emanating from A. The locus of the vertex C of the triangle is:
  1. A straight line AB, $1\frac{1}{2}$ cm from A

  2. A circle with A as centre and radius 2 cm

  3. A circle with A as centre and radius 3 cm

  4. A circle with radius 3 cm and centre 4 cm from B along BA

  5. An ellipse with A as a focus
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let A be at the origin (0,0) and B at (2,0). Median from A to BC has length 1.5. Let M be the midpoint of BC. AM = 1.5. The locus of M is a circle centered at A with radius 1.5. Since M is the midpoint of BC, C = 2M - B. This transformation maps the circle of M to a circle of C with radius 3, shifted by the vector -B.

Multiple choice

A variable chord is drawn through the origin to the circle $x^{2} + y^{2} - 2ax = 0$. The locus of the centre of the circle drawn on this chord as diameter is
  1. $x^{2} + y^{2} + ax = 0$
  2. $x^{2} + y^{2} + ay = 0$
  3. $x^{2} + y^{2} - ax = 0$
  4. $x^{2} + y^{2} - ay = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the chord be y = mx. The intersection with x^2 + y^2 - 2ax = 0 gives x^2 + m^2x^2 - 2ax = 0, so x = 2a/(1+m^2) and y = 2am/(1+m^2). The circle with this chord as diameter has center at (a/(1+m^2), am/(1+m^2)). Eliminating m leads to x^2 + y^2 - ax = 0.

Multiple choice

An equation of the circle in which the chord joining the points $(1,2)$ and $(2,-1)$ subtends an angle of $\displaystyle \frac { \pi }{ 4 } $ at any point on the circumference is
  1. ${ x }^{ 2 }+{ y }^{ 2 }-15=0$
  2. ${ x }^{ 2 }+{ y }^{ 2 }-6x-2y+5=0$
  3. ${ x }^{ 2 }+{ y }^{ 2 }+6x+2y-15=0$
  4. ${ x }^{ 2 }+{ y }^{ 2 }-2x-4y+4=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The chord joining (1,2) and (2,-1) has length sqrt((2-1)^2 + (-1-2)^2) = sqrt(1+9) = sqrt(10). The angle subtended at the circumference is pi/4, so the angle at the center is pi/2. This means the chord is a side of an inscribed square, and the radius R satisfies R^2 + R^2 = (sqrt(10))^2, so 2R^2 = 10, R^2 = 5. The center (h,k) is equidistant from (1,2) and (2,-1). Checking option B: center (3,1), R^2 = 3^2 + 1^2 - 5 = 5. This matches.

Multiple choice

The locus of the midpoint of all chords of the circle $\displaystyle x^{2}+y^{2}=25 $ which subtend right angle at the point (1, 2) is
  1. $\displaystyle x^{2}+y^{2}-x-2y-10=0 $
  2. $\displaystyle x^{2}+y^{2}+x-2y-10=0 $
  3. $\displaystyle x^{2}+y^{2}-x-3y-10=0 $
  4. $\displaystyle x^{2}+y^{2}-x+2y-10=0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the midpoint be (h, k). The chord equation is T = S1, which is hx + ky = h^2 + k^2. Since the chord subtends a right angle at (1, 2), the circle with the chord as diameter must pass through (1, 2). Using the property of chords subtending 90 degrees, the locus is found by substituting the midpoint coordinates into the circle equation geometry.