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

The area of triangle formed by the tangent, normal drawn at $(1,\sqrt 3 )$ to the circle ${x^2} + {y^2} = 4$ and positive $x$-axis, is

The equation of the tangents to the circle ${x^2} + {y^2} + 4x - 4y + 4 = 0$ which make equal intercepts on the positive coordinate axes is given by

The area of the triangle formed by the tangents from the points $(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ and the line joining their points of contact is

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

Let the normals at all the points on a given curve pass through a fixed point $(a, b) .$ If the curve passes through $(3,-3)$ and $(4,-2 \sqrt{2}),$ and given that $a-2 \sqrt{2} b=3,$ then $\left(a^{2}+b^{2}+a b\right)$ is equal to ..... .