QJEEAdv14Maeed9: JEE Advanced 2014 Mathematics - Class 11 JEEAdv14Maeed9 Solution: A circle S passes through the point (0, 1) and is orthogonal to …

Let, the equation of the required circle is

${x^2} + {y^2} + 2gx + 2fy + c = 0$ ..... (1)

Circle (I) cuts the circle ${(x - 1)^2} + {y^2} = 16$

i.e., ${x^2} + {y^2} - 2x = 15$ orthogonally

$\therefore$ $2( - g + 0) = - 15 + c$

or, $- 2g = - 15 + c$

The circle (1) also cuts the circle ${x^2} + {y^2} = 1$ orthogonally.

$\therefore$ 0 = $-$ 1 + c or, c = 1

$\therefore$ g = 7

Now, the circle (1) passes through the point (0, 1).

$\therefore$ $2f + 1 + c = 0$ or, $2f + 1 + 1 = 0$ or, f = $-$ 1

$\therefore$ the equation of the required circle is

${x^2} + {y^2} + 14x - 2y + 1 = 0$

whose centre is ( $-$ 7, 1) and radius $= \sqrt {49 + 1 - 1} = 7$ units

Therefore, (B) and (C) are the correct option.

Note :

The condition of the circle ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ cuts orthogonally to the circle ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ is $2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$