Q: The equation of the circle passing through (0, 1) and touching the curve y = x^{2} at (2 , 4) is ……

Solution : The equation of tangent to the curve y = x^{2} at (2, 4) is 4x –y = 4.

Family of circles touching the curve y = x^{2} at (2, 4) is

(x – 2)^{2} + ( y – 4)^{2} + λ(4x – y –4) = 0.

Since the required circle passes through (0 , 1), we have

4 + 9 – 5 λ = 0 ⇒ λ = 13/5

Hence equation of the required circle is

(x –2)^{2} + (y – 4)^{2} + (13/5)(4 x – y -4) = 0

⇒ 5x^{2} + 5y^{2} + 32 x – 53 y + 48 = 0.