Find the centre and radius of the circle $x^{2}+y^{2}-4x-8y-45=0$.

If the equation $x^{2}+y^{2}-10x+21=0$ has real roots $x=\alpha$ and $y=\beta,$ then

Let $\alpha, \beta$ be the roots of $x^2+5x+6=0$ and $\gamma, \delta$ be the roots of $y^2+6y+7=0$. Then the equation of the circle with $(\alpha, \gamma)$ and $(\beta, \delta)$ as the extremities of a diameter is

The circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$ also passes through the point

Let a circle $C$ have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of $C$ on the line $x + y = 1$ is $\sqrt{14}$, then the square of the radius of $C$ is . . . . . .

The equation of the circle which touches the $x$-axis at $(3, 0)$ and passes through $(1, 4)$ is given by