If a circle $S$ passing through the points $A(1, 2)$ and $B(2, 1)$ has its centre $C$ located in the third quadrant at a distance of $\frac{7}{\sqrt{2}}$ units from the line $AB$,then the point $P(1, -2)$:

If the points $(2, 0), (0, 1), (4, 5)$ and $(0, c)$ are concyclic,then $c$ is equal to

$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

Tangent $L_1 \equiv 3x - 4y - 8 = 0$ and the chord $L_2 \equiv x + y - 1 = 0$ are at a distance of $2$ and $\sqrt{2}$ units respectively from the centre of a circle $S$. $(h, k)$ is the centre of $S$ such that $h^2 + k^2 = 13$. If the midpoint of the chord $L_2 = 0$ is $(\alpha, \beta)$ and the radius of the circle is $r$,then $\alpha + \beta + r =$

If $h, k, p, q \neq 0$ and the circles $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$ touch each other at the origin,then $hq-pk-\frac{hq}{pk}$ is equal to

If the foot of the normal from the point $(4,3)$ to a circle is $(2,1)$ and $2x-y-2=0$ is a diameter of the circle,then the equation of the circle is