A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$
$(B,D)$
$(B,C)$
$(A,C)$
$(A,D)$
The intercept on the line $y = x$ by the circle ${x^2} + {y^2} - 2x = 0$ is $AB$ . Equation of the circle with $AB$ as a diameter is
The number of common tangents to the circles ${x^2} + {y^2} - x = 0,\,{x^2} + {y^2} + x = 0$ is
If a circle passes through the point $(1, 2)$ and cuts the circle ${x^2} + {y^2} = 4$ orthogonally, then the equation of the locus of its centre is
The radical centre of the circles ${x^2} + {y^2} - 16x + 60 = 0,\,{x^2} + {y^2} - 12x + 27 = 0,$ ${x^2} + {y^2} - 12y + 8 = 0$ is
$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the co-ordinate axes cut at right angles, then :