Given circles
$x^{2}+y^{2}-2 x-15=0$
$x^{2}+y^{2}-1=0$
Radical axis $x+7=0$
Centre of circle lies on (1)
Let the centre be $(-7, k )$
Let equation be $x^{2}+y^{2}+14 x-2 k y+c=0$
Orthogonallity gives
$-14= c -15 $
$\Rightarrow c =1 \,\,\,...(2)$
$(0,1) \rightarrow 1-2 k +1=0$
$\Rightarrow k =1 $
Hence radius $=\sqrt{7^{2}+ k ^{2}- c } $
$=\sqrt{49+1-1}=7$
Alternate solution
Given circles $x ^{2}+ y ^{2}-2 x -15=0$
$x^{2}+y^{2}-1=0$
Let equation of circle $x ^{2}+ y ^{2}+2 gx +2 fy + c =0$
Circle passes through $(0,1)$
$\Rightarrow 1+2 f+c=0$
Applying condition of orthogonality
$-2 g=c-15,0=c-1 $
$\Rightarrow c=1, g=7, f=-1 $
$r=\sqrt{49+1-1}=7 ; $
centre $(-7,1)$