Find the centre and radius of the circle $(x+5)^{2}+(y-3)^{2}=36$.

The equation of the circumcircle of the triangle formed by the lines $x = 0$,$y = 0$,and $2x + 3y = 5$ is

If the equation $3x^2 + (3 - p)xy + qy^2 - 2px = 8pq$ represents a circle, then the area (in sq. units) of this circle is (in $\pi$)

If a circle passing through $(1, -2)$ has $x - y = 2$ and $2x + 3y = 14$ as its diameters,then the radius of the circle is

$A$ circle cuts off positive intercepts $5$ and $6$ on the $x$ and $y$ axes respectively,and passes through the origin. Then the equation of the circle is

The equation of the circle which passes through the points $(2, 3)$ and $(4, 5)$ and whose centre lies on the straight line $4x - y - 3 = 0$ is: