Conic-Sections Question 552
Question: If from any point P, tangents PT, PT? are drawn to two given circles with centres A and B respectively; and if PN is the perpendicular form P on their radical axis, then $ PT^{2}-PT{{’}^{2}}= $
Options:
A) PN.AB
B) $ 2PN.AB $
C) $ 4PN.AB $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the two given circles be $ x^{2}+y^{2}+2g_1x+c=0 $ ? (1) And $ x^{2}+y^{2}+2g_2x+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^2_1+y^2_1+2g_1x_1+c}; $ $ PT=\sqrt{x^2_1+y^2_1+2g_2x_1+c} $ Radical axis of (1) and (2) is $ 2(g_1-g_2)x=0 $ or $ x=0, $ $ PN= $ Length of $ \bot $ from P on radical axis $ =x_1. $
$ \therefore PT^{2}-PT{{’}^{2}} $ $ =(x^2_1+y^2_1+2g_1x_1+c)-(x^2_1+y^2_1+2g_2x_1+c) $ $ =2x_1(g_1-g_2)=2PN.AB $
BETA
