The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at $(1,-2)$ 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 $........$.

If $3x + y + k = 0$ is a tangent to the circle $x^2 + y^2 = 10$,then $k = . . . . . . $.

The equation of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(a, b)$ is $ax + by - \lambda = 0$,where $\lambda$ is:

The equation of the tangent at the point $\left( \frac{ab^2}{a^2 + b^2}, \frac{a^2b}{a^2 + b^2} \right)$ to the circle $x^2 + y^2 = \frac{a^2b^2}{a^2 + b^2}$ is

The equation of the tangent to the circle,given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it,is