A circle is drawn with $y- $ axis as a tangent and its centre at the point which is the reflection of $(3, 4)$ in the line $y = x$. The equation of the circle is
$x^2 + y^2 - 6x - 8y + 16 = 0$
$x^2 + y^2 - 8x - 6y + 16 = 0$
$x^2 + y^2 - 8x - 6y + 9 = 0$
$x^2 + y^2 - 6x - 8y + 9 = 0$
The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$, then its radius is
The equation of the tangent to the circle ${x^2} + {y^2} - 2x - 4y - 4 = 0$ which is perpendicular to $3x - 4y - 1 = 0$, is
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
If the centre of a circle is $(-6, 8)$ and it passes through the origin, then equation to its tangent at the origin, is
If the centre of a circle is $(2, 3)$ and a tangent is $x + y = 1$, then the equation of this circle is