Answer:
Option B,C
Explanation:
Plan
(i) The general equation of a circle is
$x^{2}+y^{2}+2gx+2fy+c=0$
where, centre and radius are given by (-g, -f) and $\sqrt{g^{2}+f^{2}-c}$ respectively.
(ii) If the two circles
$x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and
$x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ are orthogonal , then
$2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$
Let circle be
$x^{2}+y^{2}+2gx+2fy+c=0$
It passes through (0,1)
$\therefore$ 1+2f+c=0 ........(i)
Orthogonal with
$x^{2}+y^{2}-2x-15=0$
2g(-1)=c-15
$\Rightarrow$ c=15-2g .............(ii)
Orthogonal with
$x^{2}+y^{2}-1=0$
c=1 ...........(iii)
$\Rightarrow$ g=7 and f=-1
Centre is (-g, -f) =(-7,1)
Radius= $\sqrt{g^{2}+f^{2}-c}=\sqrt{49+1-1}=7$
