JEE Advance - Mathematics (1986 - No. 19)
Lines 5x + 12y - 10 = 0 and 5x - 12y - 40 = 0 touch a circle $$C_1$$ of diameter 6. If the centre of $$C_1$$ lies in the first quadrant, find the equation of the circle $$C_2$$ which is concentric with $$C_1$$ and cuts intercepts of length 8 on these lines.
x^2 + y^2 - 10x - 4y + 4 = 0
x^2 + y^2 + 10x - 4y + 4 = 0
x^2 + y^2 - 10x + 4y + 4 = 0
x^2 + y^2 - 10x - 4y - 4 = 0
x^2 + y^2 + 10x + 4y + 4 = 0