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

The two circles which pass through $(0, a)$ and $(0, -a)$ and touch the line $y = mx + c$ will intersect each other at a right angle,if

The equations of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$. The line joining the points $\left(-\frac{22}{7}, -4\right)$ and $\left(-\frac{1}{7}, 3\right)$ intersects the circle at only one point $P(\alpha, \beta)$. Then $17\beta - \alpha$ is equal to

The square of the length of the tangent drawn from the point $(\alpha, \beta)$ to the circle $ax^2 + ay^2 = r^2$ is

If the tangent at the point $P$ on the circle $x^2+y^2+6x+6y=2$ meets the straight line $5x-2y+6=0$ at a point $Q$ on the $Y$-axis,then the length of $PQ$ is

The equations of the tangents to the circle $x^2 + y^2 = 36$ which are inclined at an angle of $45^\circ$ to the $x$-axis are