A circle S passes through the point (0, 1) and is orthogonal to the...

Given circles
${x}^{2} + {y}^{2} - 2x - 15 = 0$
${x}^{2} + {y}^{2} - 1 = 0$
Radical axis $x + 7 = 0$ ......(i)
Centre of circle lies on (i)
Let the centre be $\left(- 7, k\right)$
Let equation be ${x}^{2} + {y}^{2} + 14x - 2ky + c = 0$
Orthogonallity gives
$- 14 = c - 15 ⇒ c = 1$ .......(ii)
\left(0, 1\right) \to 1 - 2k + 1 = 0 ⇒ k = 1$<br >Hence radius$= \sqrt{{7}^{2} + {k}^{2} - c} = \sqrt{49 + 1 - 1} = 7$<br >Alternate Solution<br >Given circles${x}^{2} + {y}^{2} - 2x - 15 = 0$<br >${x}^{2} + {y}^{2} - 1 = 0$<br >Let equation of circle${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$<br >Circle passes through (0, 1) <br >$⇒ 1 + 2f + c = 0$<br >Applying condition of orthogonality<br >$- 2g = c - 15, 0 = c - 1$<br >$⇒ c = 1, g = 7, f = - 1$<br >$r = \sqrt{49 + 1 - 1} = 7;$centre \left(- 7, 1\right)$