The point $(5, -7)$ lies outside the circle:

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

If the coordinates of one end of the diameter of the circle $x^2 + y^2 - 8x - 4y + c = 0$ are $(-3, 2)$,then the coordinates of the other end are:

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

If $(a, b)$ is the midpoint of the chord $2x - y + 3 = 0$ of the circle $x^2 + y^2 + 6x - 4y + 4 = 0$,then $2a + 3b =$

Let a circle $C: (x-h)^{2} + (y-k)^{2} = r^{2}, k > 0$,touch the $x$-axis at $(1, 0)$. If the line $x + y = 0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $PQ$ is $2$,then the value of $h + k + r$ is equal to