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

Circle $K$ is inscribed in the first quadrant touching the circle $x^{2}+y^{2}=36$ internally. The length of the radius of the circle $K$ is ?
  1. $\dfrac { 6-\sqrt { 2 } }{ 2 } $
  2. $\dfrac { 3-\sqrt { 2 } }{ 2 } $
  3. $3$
  4. $6\left( \sqrt { 2 } -1 \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The circle x^2 + y^2 = 36 has radius 6 and center (0,0). A circle K with radius r in the first quadrant touching the axes and the circle internally has center (r, r). The distance from (0,0) to (r,r) is r*sqrt(2). The condition for internal tangency is r*sqrt(2) + r = 6. Thus r(sqrt(2)+1) = 6, so r = 6 / (sqrt(2)+1) = 6(sqrt(2)-1).

Multiple choice

The equation of the circle and its chord are respectively${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$ are $x cos\alpha +y sin\alpha =p$. The equation of the circle of which this chord is diameter is-
  1. ${x}^{2}+{y}^{2}-{ 2 }_{ PX }cos\alpha -{ 2 }_{ py }sin\alpha +{ 2 }_{ { p }^{ 2 } }-{ a }^{ 2 }=0$
  2. ${x}^{2}+{y}^{2}-{ 2 }_{ PX }cos\alpha -{ 2 }_{ py }sin\alpha + { p }^{ 2 } -{ a }^{ 2 }=0$
  3. ${x}^{2}+{y}^{2}-{ 2 }+{ PX }cos\alpha -{ 2 }+{ py }sin\alpha +{ 2 }_{ { p }^{ 2 } }-{ a }^{ 2 }=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Equation of the circle which is such that the lengths of the tangents to it from the points $( 1,0 ) , ( 0,2 )$ and $( 3,2 )$ are $1 , \sqrt { 7 }$ and $\sqrt { 2 }$ respectively is
  1. $6 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
  2. $9 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
  3. $3 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
  4. $x ^ { 2 } + y ^ { 2 } - 28 x - 5 y + 28 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Length of tangent from (x1, y1) to circle x^2+y^2+2gx+2fy+c=0 is sqrt(x1^2+y1^2+2gx1+2fy1+c). Square the lengths: 1 = 1+2g+c, 7 = 4+4f+c, 2 = 9+4+6g+4f+c. Solve the system: 2g+c=0, 4f+c=3, 6g+4f+c=-11. From 1st, c=-2g. From 2nd, 4f-2g=3. From 3rd, 6g+4f-2g=-11 => 4g+4f=-11. Subtract: 6g=-14 => g=-7/3. Then c=14/3. 4f=3+2g=3-14/3=-5/3 => f=-5/12. Equation: x^2+y^2-14/3x-5/6y+14/3=0. Multiply by 6: 6(x^2+y^2)-28x-5y+28=0.

Multiple choice

An equilateral triangle is inscribed in the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-12\mathrm{x}-8\mathrm{y}+4=0$ then the length of the side is
  1. $2$
  2. $6$
  3. $4\sqrt{3}$
  4. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The circle equation x^2 + y^2 - 12x - 8y + 4 = 0 can be rewritten as (x-6)^2 + (y-4)^2 = 36 + 16 - 4 = 48. The radius r is sqrt(48) = 4*sqrt(3). For an equilateral triangle inscribed in a circle, the side length s is r*sqrt(3). Thus, s = 4*sqrt(3) * sqrt(3) = 4 * 3 = 12.

Multiple choice

A line meets the co-ordinate axes in $A$ and $B$. A circle is circumscribed about the triangle $OAB$. If $d_1$ and $d_2$ are the distances of the tangent to the circle at the origin $O$ from the points $A$ and $B$, respectively, then the diameter of the circle is
  1. $\displaystyle \frac{2d_1 + d_2}{2}$
  2. $\displaystyle \frac{d_1 + 2d_2}{2}$
  3. $d_1 + d_2$
  4. $\displaystyle \frac{d_1 d_2}{d_1 + d_2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a line intercepting axes at A(a,0) and B(0,b), the circle with diameter AB has equation x^2 + y^2 - ax - by = 0. The tangent at the origin is the line passing through origin perpendicular to the radius. The distances d1 and d2 relate to the diameter such that the diameter is d1 + d2.

Multiple choice

A pair of straight lines $x^{2} - 8x + 12 = 0$ and $y^{2} - 14y + 45 = 0$ are forming a square. What is the centre of circle inscribed in the square.
  1. $(3, 2)$
  2. $(7, 4)$
  3. $(4, 7)$
  4. $(0, 1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The lines x^2 - 8x + 12 = 0 factor to (x-2)(x-6)=0, so x=2 and x=6. The lines y^2 - 14y + 45 = 0 factor to (y-5)(y-9)=0, so y=5 and y=9. These lines form a square with vertices (2,5), (6,5), (6,9), and (2,9). The center of the inscribed circle is the center of the square, which is the midpoint of the diagonals: ((2+6)/2, (5+9)/2) = (4, 7).