The equation of a circle that intersects the circle $x^2 + y^2 + 14x + 6y + 2 = 0$ orthogonally and whose centre is $(0, 2)$ is

Let $S \equiv x^2+y^2-6x-6y+4=0$ and $S^{\prime} \equiv x^2+y^2-2x-4y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having the same radical axis as $S=0$ and $S^{\prime}=0$ is

From a point $P$ on the circle $x^2+y^2-4x-6y+9=0$,a pair of tangents $PQ$ and $PR$ are drawn touching the circle $x^2+y^2-4x-6y+12=0$ at $Q$ and $R$. If $C$ is the centre of the concentric circles,then the area of the $\triangle CQR$ (in sq. units) is

The centre of a circle which cuts $x^{2}+y^{2}+6x-1=0$,$x^{2}+y^{2}-3y+2=0$ and $x^{2}+y^{2}+x+y-3=0$ orthogonally is

If the circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ touch each other internally,then $\alpha$ is equal to

$(a, 0)$ and $(b, 0)$ are the centres 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