Let $S$ be the focus of the parabola $y^2=8x$ and let $PQ$ be the common chord of the circle $x^2+y^2-2x-4y=0$ and the given parabola. The area of the triangle $PQS$ is:

If the point of intersection of the tangents drawn at the points where the line $5x + y + 1 = 0$ cuts the circle $x^2 + y^2 - 2x - 6y - 8 = 0$ is $(a, b)$,then $5a + b =$

If the two circles $x^2+y^2-2x-6y+10-r^2=0$ and $x^2+y^2-8x+2y+8=0$ have a common chord of non-zero length,then

The length of the common chord of the circles $x^2 + y^2 - 6x - 16 = 0$ and $x^2 + y^2 - 8y - 9 = 0$ is:

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then $(4g-3)(f-2)=$

$C_1$ is the circle with centre at $O(0,0)$ and radius $4$,$C_2$ is a variable circle with centre at $(\alpha, \beta)$ and radius $5$. If the common chord of $C_1$ and $C_2$ has slope $\frac{3}{4}$ and is of maximum length,then one of the possible values of $\alpha+\beta$ is