Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6 x-4 y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of the triangle $P A B$ is

The equations of tangents to the circle ${x^2} + {y^2} - 22x - 4y + 25 = 0$ which are perpendicular to the line $5x + 12y + 8 = 0$ are

Let the lengths of intercepts on $x$ -axis and $y$ -axis made by the circle $x^{2}+y^{2}+a x+2 a y+c=0$ $(a < 0)$ be $2 \sqrt{2}$ and $2 \sqrt{5}$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x +2 y =0,$ is euqal to :

The co-ordinates of the point from where the tangents are drawn to the circles ${x^2} + {y^2} = 1$, ${x^2} + {y^2} + 8x + 15 = 0$ and ${x^2} + {y^2} + 10y + 24 = 0$ are of same length, are

The line $2 x - y +1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2 y=4$. Then, the radius of the circle is

Let the lines $y+2 x=\sqrt{11}+7 \sqrt{7}$ and $2 y + x =2 \sqrt{11}+6 \sqrt{7}$ be normal to a circle $C:(x-h)^{2}+(y-k)^{2}=r^{2}$. If the line $\sqrt{11} y -3 x =\frac{5 \sqrt{77}}{3}+11$ is tangent to the circle $C$, then the value of $(5 h-8 k)^{2}+5 r^{2}$ is equal to.......