The equation of the normal to the circle $2x^2 + 2y^2 - 2x - 5y + 3 = 0$ at the point $(1, 1)$ is:

The square of the length of the tangent from $(3, -4)$ to the circle $x^2 + y^2 - 4x - 6y + 3 = 0$ 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 line $3x - 4y = 0$ is:

Let the straight line $y=2x$ touch a circle with center $(0, \alpha)$,$\alpha>0$,and radius $r$ at a point $A_1$. Let $B_1$ be the point on the circle such that the line segment $A_1 B_1$ is a diameter of the circle. Let $\alpha+r=5+\sqrt{5}$. Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P) \alpha \text{ equals}$	$(1) (-2,4)$
$(Q) r \text{ equals}$	$(2) \sqrt{5}$
$(R) A_1 \text{ equals}$	$(3) (-2,6)$
$(S) B_1 \text{ equals}$	$(4) 5$
	$(5) (2,4)$

Find the equation of the normal to the circle $x^2 + y^2 - 5x + 2y - 48 = 0$ at the point $(5, 6)$.