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
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5840496783781950270)
∴ 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