Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6 x+4 y+8=0$ at the point $P$ and $Q$ on it. If the circumcircle of the triangle OPQ passes through the point $\left(\alpha, \frac{1}{2}\right)$, then a value of $\alpha$ is

Two tangents are drawn from a point $P$ to the circle $x^{2}+y^{2}-2 x-4 y+4=0$, such that the angle between these tangents is $\tan ^{-1}\left(\frac{12}{5}\right)$, where $\tan ^{-1}\left(\frac{12}{5}\right) \in(0, \pi)$. If the centre of the circle is denoted by $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\Delta PAB$ and $\Delta CAB$ is :

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 at $P$, any point on the circle ${x^2} + {y^2} = 4$, meets the coordinate axes in $A$ and $B$, then

Equation of the tangent to the circle, at the point $(1 , -1)$ whose centre is the point of intersection of the straight lines $x - y = 1$ and $2x + y= 3$ is

Given the circles ${x^2} + {y^2} - 4x - 5 = 0$and ${x^2} + {y^2} + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha ,\beta )$such that the tangents from P to both the circles are equal, then