JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Circle

Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. The given circles are $x^2+y^2-2x-15=0$ and $x^2+y^2-1=0$. For orthogonality with the first circle: $2(-1)(g)+2(0)(f)=-15+c$, giving $-2g=-15+c$. For orthogonality with the second circle: $2(0)(g)+2(0)(f)=-1+c$, giving $c=1$. From these, $g=7$. Since $S$ passes through $(0,1)$: $0+1+0+2f+1=0$, giving $f=-1$. So the circle is $x^2+y^2+14x-2y+1=0$, with centre $(-7,1)$ and radius $\sqrt{49+1-1}=7$. Thus (B) and (C) are correct.