Question

Find the equation of the circle with radius 5, whose centre lies on x-axis and passes through the point (2, 3).

Answer 1

r = 5 . 

Centre lies on X – axis ⇒ pt(2, 3) = (x, y) 

Let the circle equation be 

x² + y² + 2fx + 2gx + C = 0 … (1) 

∵ centre lies on x-axis ∴ f = 0 

∴ (1) becomes x² + y² + 2gx + c = 0) at (2,3) 

4 + 9 + 2g(2) + c = 0 ∴ 48 + c = -13 …(2) 

Given radius = 5

∴ g² + 4g – 12 = 0 ⇒ g² + 5g – 2y = 12 = 0 

⇒ g(g – 16) – 2(g + 6)=0 

∴ g = -6 or g = -2 

Substitute in (2) 4g + C = -13 

at g = -6 4(-6) + C = -13 ⇒ C = -13 + 24 = 11 

at g = -2 4(-2) + 1 = -13 ⇒ C =-13 + 8 = -5 

∴ at g = -6, C = 11, f = 0 (1) becomes 

x² + y² – 12x + 11 = 0

at g = -2, c = -5, f = 0 

x² + y² – 4x – 5 = 0