The equation of circle with centre $(1, 2)$ and tangent $x + y - 5 = 0$ is
${x^2} + {y^2} + 2x - 4y + 6 = 0$
${x^2} + {y^2} - 2x - 4y + 3 = 0$
${x^2} + {y^2} - 2x + 4y + 8 = 0$
${x^2} + {y^2} - 2x - 4y + 8 = 0$
If the line $3x - 4y = \lambda $ touches the circle ${x^2} + {y^2} - 4x - 8y - 5 = 0$, then $\lambda $ is equal to
If the line $y = mx + c$be a tangent to the circle ${x^2} + {y^2} = {a^2}$, then the point of contact is
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 two circles which passes through $(0,a)$ and $(0, - a)$ and touch the line $y = mx + c$ will intersect each other at right angle, if
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