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

AB is chord of the circle $x^2+y^2=25$. The tangents of A and B intersect at C. If $(2,3)$ is the mid-point of AB, then area of the quadrilateral OACB is
  1. $50\sqrt { \frac { 13 }{ 3 } } $
  2. $50\sqrt { \frac { 3 }{ 13 } } $
  3. $50\sqrt3$
  4. $\dfrac{50}{\sqrt3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The chord AB has midpoint (2,3). The line from the origin to the midpoint is perpendicular to the chord. The area of quadrilateral OACB is 2 * Area(triangle OAC). Area(OAC) = 1/2 * base * height. This is a standard geometry problem for a circle.

Multiple choice

Find the equation of the circle which passes through the origin and the cuts off chords of the lengths $4$ and $6$ on the positive side of $x-axis$ and $y-axis$ respectively.
  1. $x^{2}+y^{2}-4x-6y=0$
  2. $x^{2}-y^{2}+4x+6y=0$
  3. $-x^{2}+y^{2}+4x-6y=0$

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

Circle passes through (0,0). Chords on axes mean the circle intersects axes at (4,0) and (0,6). The equation of a circle passing through (0,0), (a,0), and (0,b) is x^2 + y^2 - ax - by = 0. Here a=4, b=6, so x^2 + y^2 - 4x - 6y = 0.

Multiple choice

Consider a family of circles which are passing through the point $\left ( -1,1 \right )$ and are tangent to $x$-axis. If $\left ( h,k \right )$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval
  1. $\displaystyle -\frac{1}{2}\leq k\leq \displaystyle \frac{1}{2}$
  2. $k\leq \displaystyle \frac{1}{2}$
  3. $0\leq k\leq \displaystyle \frac{1}{2}$
  4. $k\geq \displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A circle passing through (-1, 1) and tangent to the x-axis has a radius r = |k|. The equation is (x-h)^2 + (y-k)^2 = k^2. Substituting (-1, 1): (-1-h)^2 + (1-k)^2 = k^2. Expanding: 1 + 2h + h^2 + 1 - 2k + k^2 = k^2, so h^2 + 2h + 2 = 2k. Since h^2 + 2h + 2 = (h+1)^2 + 1, the minimum value is 1. Thus 2k >= 1, or k >= 1/2.

Multiple choice

If chord PQ of a circle have length equal to the radius then the distance of chord from centre in terms of radius is
  1. $\displaystyle \frac{\sqrt{3}}{4} r$
  2. $\displaystyle \frac{\sqrt{5}}{2} r$
  3. $ \frac { 3\sqrt { 3 } }{ 2 } r $
  4. $\displaystyle \frac{\sqrt{3}}{2} r$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If the chord length equals the radius, it forms an equilateral triangle with the two radii connecting the center to the chord endpoints. The distance from the center to the chord is the altitude of this equilateral triangle, which is calculated as r * sin(60 degrees) = r * sqrt(3)/2.

Multiple choice

The equation of the circle which passes through origin and cuts off chords of length 2 on the lines $x = y$ and $x = -y$ is
  1. $\mathrm{x}^{2}+\mathrm{y}^{2}\pm 2\sqrt{2}\mathrm{y}=0,\ \mathrm{x}^{2}+\mathrm{y}^{2}\pm 2\sqrt{2}\mathrm{x}= 0$
  2. $\mathrm{x}^{2}+\mathrm{y}^{2}\pm 3\sqrt{3}\mathrm{x}=0,\ \mathrm{x}^{2}+\mathrm{y}^{2}$ $\pm 3\sqrt{3}\mathrm{y}=0$
  3. $\mathrm{x}^{2}+\mathrm{y}^{2}\pm 4\sqrt{2}\mathrm{x}=0,\ \mathrm{x}^{2}+\mathrm{y}^{2}\pm 4\sqrt{2}\mathrm{y}= 0$
  4. $\mathrm{x}^{2}+\mathrm{y}^{2}\pm 4\sqrt{2}\mathrm{x}=0,\ \mathrm{x}^{2}+\mathrm{y}^{2}\pm 4\sqrt{3}\mathrm{y}= 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle passes through the origin, so its equation is x^2 + y^2 + 2gx + 2fy = 0. The chord length formula for a line ax + by + c = 0 is 2 * sqrt(r^2 - d^2). Setting chord length to 2 for lines x=y and x=-y allows solving for g and f.

Multiple choice

The locus of the centre of a circle which passes through the point $(h,k)$ and cuts off a chord of length $2d$ on the line $lx + my + n = 0$ is
  1. $(l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n})^{2}=(l^{2}+\mathrm{m}^{2})[(x-h)^{2}+(y-k)^{2}-d^{2}]$
  2. $(l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n})^{2}=(l^{2}+\mathrm{m}^{2}) [(y-h)^{2}+(x-k)^{2}-d^{2}]$
  3. $(l\mathrm{x}+$ my $+\mathrm{n})^{2}=\mathrm(l^{3}+\mathrm{m}^{3})\lfloor(2x-2h)^{2}+(2y-10k)^{2}-d^{2}\rfloor$
  4. $(l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n})^{2}=(l^{3}+\mathrm{m}^{3})[(2x-2h)^{2}+(2y-10k)^{2}-d^{2}]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a circle with centre (x,y) passing through (h,k), the radius squared is (x-h)^2 + (y-k)^2. A chord of length 2d has distance squared r^2 - d^2 from the centre to its line, which gives the equation in option A.

Multiple choice

The equation of the locus of the middle point of a chord of the circle ${ x }^{ 2 }+{ y }^{ 2 }=2(x+y)$ such that the point of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the x-axis is
  1. $x + y = 2$
  2. $x - y = 2$
  3. $2x - y = 1$

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

The locus of the midpoint of a chord is found by using the condition that the chord is bisected at the point (h, k). The condition of equal inclination to the x-axis for the lines joining the origin to the intersection points leads to the equation x + y = 2.

Multiple choice

A circle touches the sides AB and AD of the rectangle ABCD at P and Q respectively and passes through the vertex C. If the distance of C from chord PQ is 5 units and the area of the rectangle is $\lambda$, then find the value of $\frac{\lambda }{5}$

  1. 2

  2. 5

  3. 10

  4. 25
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the circle have radius R. The rectangle has sides a and b. The distance from C to chord PQ is related to the geometry of the circle. Based on standard problems of this type, the area lambda is 25, so lambda/5 = 5.

Multiple choice

The line $3x+5y+9=0$ w.r.t the circle $x^{2}+y^{2}-4x+6y+5=0$ is
  1. chord dividing circumference in $1:3$ ratio

  2. diameter

  3. tangent

  4. outside line
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The center of the circle x^2 + y^2 - 4x + 6y + 5 = 0 is (2, -3). Substituting these coordinates into the line equation 3x + 5y + 9 = 0 gives 3(2) + 5(-3) + 9 = 6 - 15 + 9 = 0. Since the center lies on the line, the line is a diameter.