Given two circles $x^2+y^2+8x-6y-24=0$ and $x^2+y^2-4x+10y+20=0$. Then they are

$A$ rectangle is inscribed in a circle with a diameter lying along the line $3y = x + 7$. If the two adjacent vertices of the rectangle are $(-8, 5)$ and $(6, 5)$,then the area of the rectangle (in $sq. units$) is

The line $2x - 3y = 1$ divides the circular region $x^2 + y^2 \leq 6$ into two parts. If $S = \left\{ \left(2, \frac{3}{4}\right), \left(\frac{5}{2}, \frac{3}{4}\right), \left(\frac{1}{4}, -\frac{1}{4}\right), \left(\frac{1}{8}, \frac{1}{4}\right) \right\}$,then the number of points in the set $S$ that lie inside the smaller part is:

Let $PA$ and $PB$ be the tangent segments drawn from point $P(6, 8)$ to the circle with the centre at origin $O(0, 0)$. The radius $r$ of the circle for which the area of quadrilateral $PAOB$ is maximum, is...

The circle touching the coordinate axes with its centre lying on $x-2y-3=0$ is

Three circles of radius $r$ touch each other externally. What is the radius of the circle that touches all three circles internally?