Question

If from any point $P$, tangents $PT,P{T}^{\prime }$ are drawn to two given circles with cenetres $A$ and $B$, respectively ; and if $PN$ is the perpendicular from $P$ on their radical axis, then ${PT}^2-{P{T}^{\prime }}^2=$

• A. $2.PN.AB$
• B. $4PN.AB$
• C. $PN.AB$
• D. None of these

Answer 1

The correct option is A $2.PN.AB$

Let the two given circles be $x^2+y^2+2g_{1}x+c=0$ ...(1)

and $x^2+y^2+2g_{2}x+c=0$ ...(2)

Their centres are $A(-g_1,0)$ and $B(-g_2,0).$

$\therefore AB=g_1-g_2.$

Let $P$ be the point $(x_1,y_1).$

Then, $PT=\sqrt{x_1^2+y_1^2+2g_1{x}_1+c;}$

$P{T}^{\prime }=\sqrt{x_1^2+y_1^2+2g_2{x}_1+c}$

Radical axis of (1) and (2) is $2(g_1-g_2)x=0$ or $x=0.$

$PN=$length of $\perp$ from $P$ on radical axis $x_1.$

$\therefore {PT}^2-{PT}^{\prime 2}$

$=(x_1^2+y_1^2+2g_1{x}_1+c)-(x_1^2+y_1^2+2g_2{x}_1+c)$

$=2x_1(g_1-g_2)=2PN.AB$

$\therefore k=2$