Eliminating y, we get
(3c22 c - 2)x = c2 - 1 ∴x=(c−1)(c+1)(c−1)(3c+2)=c+13c+2
∴ limit when c ⇒ 1 gives x=25 and hence y=−125.
∴ Centre ( g , f) is (25,125).
Circle is
x2+y2−(45)x+(225)y+k=0
It passes through (2,0) ∴ k=−(125)
Hence the required circle is
25(x2+y2)−20x+60y=0.