JEE Main 10 Jan 2019 Shift 1 Solved Paper

x2+y2 + 4x – 6y – 12 = 0
Equation of tangent at (1, –1)
x – y + 2(x + 1) – 3(y – 1) – 12 = 0
3x – 4y – 7 = 0
∴ Equation of circle is
(x2+y2 + 4x – 6y – 12) + λ(3x – 4y – 7) = 0
It passes through (4, 0) :
(16 + 16 – 12) + λ(12 – 7) = 0
⇒ 20 + λ(5) = 0
⇒ λ = –4
∴ (x2+y2 + 4x – 6y – 12) – 4(3x – 4y – 7) = 0
or x2+y2 – 8x + 10y + 16 = 0
Radius = √16+25−16 = 5