When are the tangents drawn from the origin to the circle $x^2 + 2px + y^2 - 2qy + q^2 = 0$ perpendicular to each other?

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 $List-II$.

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

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6x+4y+8=0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $OPQ$ passes through the point $(\alpha, \frac{1}{2})$,then a value of $\alpha$ is

At which point on the $y$-axis is the line $x = 0$ a tangent to the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

If the line $y = mx + c$ is a tangent to the circle $x^2 + y^2 = a^2$,then the point of contact is

Let the lengths of intercepts on the $x$-axis and $y$-axis made by the circle $x^{2}+y^{2}+ax+2ay+c=0$ $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$,respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$ is equal to: