Mathematics - Circle Question with Solution | TestHub

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
$- 14=c-15 \Rightarrow c=1$.......(ii)
$\left(0, 1\right)\to 1-2k+1=0 \Rightarrow k=1$
Hence radius$= \sqrt{7^2+k^2-c}= \sqrt{49+1-1}=7$
Alternate Solution
Given circles$x^2+y^2-2x-15=0$
$x^2+y^2-1=0$
Let equation of circle$x^2+y^2+2gx+2fy+c=0$
Circle passes through (0, 1)
$\Rightarrow$$1+2f+c=0$
Applying condition of orthogonality
$-2g=c-15, 0=c-1$
$\Rightarrow$$c=1, g=7, f= -1$
$r= \sqrt{49+1-1}=7;$centre$\left(-7, 1\right)$