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NECO 2023 · Paper 1 · Q29

Find the equation of the circle which has its centre at (−2,4)(-2, 4) and passes through (1,5)(1, 5).

Worked solution (try it first)
  1. The radius is the distance from the centre (−2,4)(-2, 4) to (1,5)(1, 5): r2=(1+2)2+(5−4)2=10r^2 = (1 + 2)^2 + (5 - 4)^2 = 10.
  2. A centre (a,b)(a, b) gives (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2, so here (x+2)2+(y−4)2=10(x + 2)^2 + (y - 4)^2 = 10.
  3. Expand: x2+4x+4+y2−8y+16=10x^2 + 4x + 4 + y^2 - 8y + 16 = 10.
  4. Take 10 from both sides: x2+y2+4x−8y+10=0x^2 + y^2 + 4x - 8y + 10 = 0, option E.

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