Find the equation of the normal to the circle $x^2 + y^2 - 5x + 2y - 48 = 0$ at the point $(5, 6)$.

The line $y = x + c$ will intersect the circle $x^2 + y^2 = 1$ in two coincident points,if

If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^{2}+y^{2}-2x-4y=0$ intersect at the point $Q$,then the area of the triangle $OPQ$ is equal to

The angle between the tangents drawn from the point $(2, 2)$ to the circle $x^2 + y^2 + 4x + 4y + c = 0$ is $\cos^{-1}\left(\frac{7}{16}\right)$. If two such circles exist,then the sum of the values of $c$ is:

$A$ triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2x$ and $x^2+y^2=4x$,and the line $x+y+2=0$. If $r$ is the radius of its circumcircle,then $r^2$ is equal to $........$.

The diagonals of a rectangle have endpoints at $(0, 0)$ and $(8, 6)$. The equations of the tangents to the circumcircle of the rectangle,which are parallel to these diagonals,are: