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

The equation of tangent to the circle ${x^2} + {y^2} = {a^2}$ parallel to $y = mx + c$ is

The tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be -

Two circles each of radius $5\, units$ touch each other at the point $(1,2)$. If the equation of their common tangent is $4 \mathrm{x}+3 \mathrm{y}=10$, and $\mathrm{C}_{1}(\alpha, \beta)$ and $\mathrm{C}_{2}(\gamma, \delta)$, $\mathrm{C}_{1} \neq \mathrm{C}_{2}$ are their centres, then $|(\alpha+\beta)(\gamma+\delta)|$ is equal to .... .

A tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The co-ordinates of another point $B$ on the circle such that $l\, (AB) = 4$ are :

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y -4=0$ is-