Let $P$ and $Q$ be the inverse points with respect to the circle $S \equiv x^2+y^2-4x-6y+k=0$ and $C$ be the centre of the circle $S=0$ such that $CP \cdot CQ=4$. If $P=(1,2)$ and $Q=(a, b)$,then $2a=$

$A$ triangle $PQR$ is inscribed in the circle $x^2 + y^2 = 25$. If the coordinates of $Q$ and $R$ are $(3, 4)$ and $(-4, 3)$ respectively,then $\angle QPR = \dots$

The square of the length of the tangent from the point $(3, -4)$ to the circle $x^2 + y^2 - 4x - 6y + 3 = 0$ is ....

$A$ straight line meets the coordinate axes at $A$ and $B$. $A$ circle is circumscribed about the triangle $OAB$, where $O$ is the origin. If $m$ and $n$ are the distances of the tangent to the circle at the origin from the points $A$ and $B$ respectively, then the diameter of the circle is

If the equation of the circle whose radius is $\sqrt{10}$ and which touches the circle $x^2+y^2+2x+8y-23=0$ externally at the point $(1,2)$ is $x^2+y^2+ax+by+c=0$,then $|a+b+c|=$

Let $C_1$ be the circle in the third quadrant of radius $3,$ that touches both coordinate axes. Let $C_2$ be the circle with centre $(1,3)$ that touches $C_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m + n$ is equal to :