The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point

The equation of the tangent to the circle at the point $(1, -1)$,whose center is the point of intersection of the straight lines $x - y = 1$ and $2x + y = 3$,is:

If the straight line $y = mx + c$ touches the circle ${x^2} + {y^2} - 4y = 0$,then the value of $c$ will be

$A$ tangent is drawn to the circle $2x^{2} + 2y^{2} - 3x + 4y = 0$ at point $A$ and it meets the line $x + y = 3$ at $B(2, 1)$. Then,the length of $AB$ is equal to:

If the tangent at the point $P$ on the circle $x^2+y^2+6x+6y=2$ meets the straight line $5x-2y+6=0$ at a point $Q$ on the $Y$-axis,then the length of $PQ$ is

$A$ is the centre of the circle $x^2+y^2-2x-4y-20=0$. If the tangents drawn at the points $B(1,7)$ and $D(4,-2)$ on the circle meet at the point $C$,then the area of the quadrilateral $ABCD$ (in square units) is