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:

If the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 2gx + g^2 - b^2 = 0$ touch each other externally,then:

The number of common tangents to the circles $x^{2}+y^{2}-y=0$ and $x^{2}+y^{2}+y=0$ is

Two circles each of radius $5 \text{ units}$ touch each other at the point $(1, 2)$. If the equation of their common tangent is $4x + 3y = 10$,and $C_{1}(\alpha, \beta)$ and $C_{2}(\gamma, \delta)$,$C_{1} \neq C_{2}$ are their centres,then $|(\alpha + \beta)(\gamma + \delta)|$ is equal to .... .

Let $S_1, S_2,$ and $S_3$ be three circles of unit radius which touch each other externally. The common tangents to each pair of circles are drawn and extended so that they intersect and form a triangle $ABC$ with circumradius $R$. Then $R$ is equal to

If two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect in two distinct points,then