Equations of Circles

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

1. CSCA Math December 2025 Past Paper
   A circle has center $(-3,2)$ and radius 4. Its equation is ( ).
   • A．$(x+3)^2+(y-2)^2=16$
   • B．$(x-3)^2+(y+2)^2=16$
   • C．$(x-3)^2+(y+2)^2=4$
   • D．$(x+3)^2+(y-2)^2=4$
   Show answer & solution

   Answer: A

   Answer: A A circle with center $(-3,2)$ and radius 4 has equation $(x+3)^2+(y-2)^2=16$. Choose A.
   See it in the full CSCA Math December 2025 Past Paper
2. CSCA Math April 2026 Past Paper
   Equation of a circle with center ( $- 1,2$ ) and radius 3 is
   • A．$x ^ { 2 } + y ^ { 2 } - 2 x - 4 y - 4 = 0$
   • B．$x ^ { 2 } + y ^ { 2 } + 2 x + 4 y - 4 = 0$
   • C．$x ^ { 2 } + y ^ { 2 } + 2 x - 4 y + 4 = 0$
   • D．$x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 4 = 0$
   Show answer & solution

   Answer: D

   Standard form: $( x + 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 9$. Expanding: $x ^ { 2 } + 2 x + 1 + y ^ { 2 } - 4 y + 4 = 9 \Rightarrow x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 4 = 0$.
   Take the full CSCA Math April 2026 Past Paper — timed & graded
3. CSCA Math June 2026 Past Paper
   The equation of the circle centered at $(2, 5)$ with a radius of 5 is
   • A．$(x-2)^2 + (y-5)^2 = 5$
   • B．$(x-2)^2 + (y-5)^2 = 25$
   • C．$(x+2)^2 + (y+5)^2 = 25$
   • D．$(x-5)^2 + (y-2)^2 = 25$
   Show answer & solution

   Answer: B

   The standard equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$. Substituting $a = 2$, $b = 5$, $r = 5$ gives $(x-2)^2 + (y-5)^2 = 25$. Choose (B).
   Take the full CSCA Math June 2026 Past Paper — timed & graded