If from any point $$P,$$ tangents $$PT,\,PT'$$   are drawn to two given circles with centers $$A$$ and $$B$$ respectively; and if $$PN$$  is the perpendicular from $$P$$ on their radical axis, then $$P{T^2} - PT{'^2} = ?$$

Solution :
Let the two given circles be
$$\eqalign{ & {x^2} + {y^2} + 2{g_1}x + c = 0......\left( 1 \right) \cr & {\text{and }}{x^2} + {y^2} + 2{g_2}x + c = 0......\left( 2 \right) \cr} $$
Their centres are $$A\left( { - {g_1},\,0} \right)$$   and $$B\left( { - {g_2},\,0} \right)$$
$$\therefore \,AB = {g_1} - {g_2}$$
Let $$P$$ be the point $$\left( {{x_1},\,{y_1}} \right).$$  Then,
$$\eqalign{ & PT = \sqrt {x_1^2 + y_1^2 + 2{g_1}{x_1} + c} \,; \cr & PT = \sqrt {x_1^2 + y_1^2 + 2{g_2}{x_1} + c} \cr} $$
Radical axis of $$\left( 1 \right)$$ and $$\left( 2 \right)$$ is $$2\left( {{g_1} - {g_2}} \right)x = 0{\text{ or }}x = 0,$$
$$PN = $$  length of $$ \bot $$ from $$P$$ on radical axis $$ = {x_1}$$
$$\eqalign{ & \therefore \,P{T^2} - PT{'^2} \cr & = \left( {x_1^2 + y_1^2 + 2{g_1}{x_1} + c} \right) - \left( {x_1^2 + y_1^2 + 2{g_2}{x_1} + c} \right) \cr & = 2{x_1}\left( {{g_1} - {g_2}} \right) \cr & = 2PN.AB \cr} $$