The equations of the tangents to the circle ${x^2} + {y^2} = 50$ at the points where the line $x + 7 = 0$ meets it, are
$7x \pm y + 50 = 0$
$7x \pm y - 5 = 0$
$y \pm 7x + 5 = 0$
$y \pm 7x - 5 = 0$
A pair of tangents are drawn from the origin to the circle ${x^2} + {y^2} + 20(x + y) + 20 = 0$. The equation of the pair of tangents is
If line $ax + by = 0$ touches ${x^2} + {y^2} + 2x + 4y = 0$ and is a normal to the circle ${x^2} + {y^2} - 4x + 2y - 3 = 0$, then value of $(a,b)$ will be
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
Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$, where $\alpha, \beta>0$. If the circle touches the line $y+x=0$ at the point $P$, whose distance from the origin is $4 \sqrt{2}$ , then $(\alpha+\beta)^2$ is equal to................
The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y -4=0$ is-