The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ lies on the line

The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle $x^2+y^2=9$ is

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

If the circles $x^{2}+y^{2}+6x+8y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y = k+6\sqrt{3}+8\sqrt{6}$ with $k>0$ touch internally at the point $P(\alpha, \beta)$,then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

If the equation of the circle of radius $3$ units which touches the circle $x^2+y^2+6x-8y-11=0$ externally at $(3,0)$ is $x^2+y^2+2gx+2fy+c=0$,then $3g-4f+c=$