Let $B$ be the centre of the circle $x^{2}+y^{2}-2 x+4 y+1=0$ Let the tangents at two points $\mathrm{P}$ and $\mathrm{Q}$ on the circle intersect at the point $\mathrm{A}(3,1)$. Then $8.$ $\left(\frac{\text { area } \triangle \mathrm{APQ}}{\text { area } \triangle \mathrm{BPQ}}\right)$ is equal to .... .

The tangent and the normal lines at the point $(\sqrt 3,1)$ to the circle $x^2 + y^2 = 4$ and the $x -$ axis form a triangle. The area of this triangle (in square units) is

If the straight line $ax + by = 2;a,b \ne 0$ touches the circle ${x^2} + {y^2} - 2x = 3$ and is normal to the circle ${x^2} + {y^2} - 4y = 6$, then the values of a and b are respectively

Tangents are drawn from $(4, 4) $ to the circle $x^2 + y^2 - 2x - 2y - 7 = 0$ to meet the circle at $A$ and $B$. The length of the chord $AB $ is

The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals $\&$ passing through the point $(2 , 1)$ is :

Equation of the pair of tangents drawn from the origin to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is