Circle And System Of Circles Question 318
Question: The area of the triangle formed by the tangents from the points (h, k) to the circle $ x^{2}+y^{2}=a^{2} $ and the line joining their points of contact is
[MNR 1980]
Options:
A) $ a\frac{{{(h^{2}+k^{2}-a^{2})}^{3/2}}}{h^{2}+k^{2}} $
B) $ a\frac{{{(h^{2}+k^{2}-a^{2})}^{1/2}}}{h^{2}+k^{2}} $
C) $ \frac{{{(h^{2}+k^{2}-a^{2})}^{3/2}}}{h^{2}+k^{2}} $
D) $ \frac{{{(h^{2}+k^{2}-a^{2})}^{1/2}}}{h^{2}+k^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of chord of contact AB is $ xh+yk=a^{2} $ …..(i)
$ OM= $ length of perpendicular from O(0, 0) on line (i) $ =\frac{a^{2}}{\sqrt{h^{2}+k^{2}}} $
$ \therefore $ $ AB=2AM=2\sqrt{OA^{2}-OM^{2}}=\frac{2a\sqrt{h^{2}+k^{2}-a^{2}}}{\sqrt{h^{2}+k^{2}}} $
Also $ PM= $ length of perpendicular from $ P(h,\ k) $ to the line (i) is $ \frac{h^{2}+k^{2}-a^{2}}{\sqrt{h^{2}+k^{2}}} $
Therefore, the required area of triangle PAB $ =\frac{1}{2}.\ AB\ .\ PM=\frac{a{{(h^{2}+k^{2}-a^{2})}^{3/2}}}{h^{2}+k^{2}} $ .
BETA
