The equation of the tangent to the circle $x^2 + y^2 - 2x - 4y - 4 = 0$ which is perpendicular to $3x - 4y - 1 = 0$ is:

The line $lx + my + n = 0$ is normal to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,if

The circle $x=5 \cos \theta, y=5 \sin \theta$ is bounded by the rectangle formed by the lines $x \pm 6=0$ and $y \pm 6=0$. The area of the triangle that lies inside the rectangle,which is formed by the tangent at $P\left(\frac{2 \pi}{3}\right)$ to the circle with two of the above given lines,is

If the tangents $x+y+k=0$ and $x+ay+b=0$ drawn to the circle $S \equiv x^2+y^2+2x-2y+1=0$ are perpendicular to each other and $k, b$ are both greater than $1$,then $b-k=$

If two tangents are drawn from the point $P$ on the circle $x^2+y^2=4$ to the circle $x^2+y^2=1$,where the point $P$ is given by $(\sqrt{2}, \sqrt{2})$,then the slopes of the tangents are:

Let the centre of a circle $C$ be $(\alpha, \beta)$ and its radius $r < 8$. Let $3x + 4y = 24$ and $3x - 4y = 32$ be two tangents and $4x + 3y = 1$ be a normal to $C$. Then $(\alpha - \beta + r)$ is equal to $........$.