Equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which is perpendicular to the straight line $y = mx + c$  is

The line $(x - a)\cos \alpha + (y - b)$ $\sin \alpha = r$ will be a tangent to the circle ${(x - a)^2} + {(y - b)^2} = {r^2}$

Let the tangent to the circle $x^{2}+y^{2}=25$ at the point $R (3,4)$ meet $x$ -axis and $y$ -axis at point $P$ and $Q$, respectively. If $r$ is the radius of the circle passing through the origin $O$ and having centre at the incentre of the triangle $OPQ ,$ then $r ^{2}$ is equal to

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.......

Line $y = x + a\sqrt 2 $ is a tangent to the circle ${x^2} + {y^2} = {a^2}$ at

If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to a circle, then the radius of the circle is