The two circles $x^2 + y^2 - 4y = 0$ and $x^2 + y^2 - 8y = 0$:

In a $\triangle ABC$,a point $D$ is chosen on $BC$ such that $BD:DC = 2:5$. Let $P$ be a point on the circumcircle of $\triangle ABC$ such that $\angle PDB = \angle BAC$. Then $PD:PC$ is:

Given two circles $x^2+y^2+8x-6y-24=0$ and $x^2+y^2-4x+10y+20=0$. Then they are

The power of a point $(2,0)$ with respect to a circle $S$ is $-4$ and the length of the tangent drawn from the point $(1,1)$ to $S$ is $2$. If the circle $S$ passes through the point $(-1,-1)$,then the radius of the circle $S$ is

If the line $x + 2by + 7 = 0$ is a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$,then $b = $

Match the statements in Column $I$ with the properties in Column $II$.

Column $I$	Column $II$
$(A)$ Two intersecting circles	$(p)$ have a common tangent
$(B)$ Two mutually external circles	$(q)$ have a common normal
$(C)$ Two circles,one strictly inside the other	$(r)$ do not have a common tangent
$(D)$ Two branches of a hyperbola	$(s)$ do not have a common normal