$y - x + 3 = 0$ is the equation of normal at $\left( {3 + \frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)$ to which of the following circles
${\left( {x - 3 - \frac{3}{{\sqrt 2 }}} \right)^2} + {\left( {y - \frac{{\sqrt 3 }}{2}} \right)^2} = 9$
${\left( {x - 3 - \frac{3}{{\sqrt 2 }}} \right)^2} + {y^2} = 6$
${(x - 3)^2} + {y^2} = 9$
${(x - 3)^2} + {(y - 3)^2} = 9$
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
The equations of the tangents drawn from the point $(0, 1)$ to the circle ${x^2} + {y^2} - 2x + 4y = 0$ are
Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6 x-4 y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of the triangle $P A B$ is
An infinite number of tangents can be drawn from $(1, 2)$ to the circle ${x^2} + {y^2} - 2x - 4y + \lambda = 0$, then $\lambda = $
The angle between the tangents to the circle ${x^2} + {y^2} = 169$ at the points $(5, 12) $ and $(12, -5)$ is ............. $^o$