The equation of the normal at $(1, 1)$ to the circle $x^2 + y^2 - x - 3y - 4 = 0$ is

Tangents drawn from the origin to the circle $x^2 + y^2 - 2ax - 2by + b^2 = 0$ are perpendicular to each other,if

If a line,$y=mx+c$ is a tangent to the circle,$(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $L_{1},$ where $L_{1}$ is the tangent to the circle,$x^{2}+y^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then

If the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ cuts each of the three circles $x^2 + y^2 + 4x + 4y + 7 = 0$,$x^2 + y^2 - 4x + 4y + 7 = 0$,and $x^2 + y^2 - 4x - 4y + 7 = 0$ orthogonally,then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $S = 0$ is

If a line passing through the origin touches the circle $(x - 4)^2 + (y + 5)^2 = 25$,then its slope is:

The angle between the tangents drawn from the origin to the circle $(x-7)^2+(y+1)^2=25$ is