Number of integral points interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt {{{\left( {x + 5\sqrt 2 } \right)}^2} + {y^2}} \, - \sqrt {{{\left( {x - 5\sqrt 2 } \right)}^2} + {y^2}\,} \, = 10$ are (where integral point $(x, y)$ means $x, y  \in  I)$

The line $y = mx + c$ will be a normal to the circle with radius $r$ and centre at $(a, b)$, if

Tangents drawn from origin to the circle ${x^2} + {y^2} - 2ax - 2by + {b^2} = 0$ are perpendicular to each other, if

Tangents to a circle at points $P$ and $Q$ on the circle intersect at a point $R$. If $P Q=6$ and $P R=5$, then the radius of the circle is

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

Points $P (-3,2), Q (9,10)$ and $R (\alpha, 4)$ lie on a circle $C$ with $P R$ as its diameter. The tangents to $C$ at the points $Q$ and $R$ intersect at the point $S$. If $S$ lies on the line $2 x - ky =1$, then $k$ is equal to $.........$.