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

A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of $60^o$. The area enclosed by these tangents and the arc of the circle is

Let $O$ be the centre of the circle $x ^2+ y ^2= r ^2$, where $r >\frac{\sqrt{5}}{2}$. Suppose $P Q$ is a chord of this circle and the equation of the line passing through $P$ and $Q$ is $2 x+4 y=5$. If the centre of the circumcircle of the triangle $O P Q$ lies on the line $x+2 y=4$, then the value of $r$ is. . . .

A tangent to the circle ${x^2} + {y^2} = 5$at the point $(1,-2)$ the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$

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

A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is