Circle And System Of Circles Question 363
Question: A pair of tangents are drawn from the origin to the circle $ x^{2}+y^{2}+20(x+y)+20=0 $ . The equation of the pair of tangents is
[MP PET 1990]
Options:
A) $ x^{2}+y^{2}+10xy=0 $
B) $ x^{2}+y^{2}+5xy=0 $
C) $ 2x^{2}+2y^{2}+5xy=0 $
D) $ 2x^{2}+2y^{2}-5xy=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of pair of tangents is given by $ SS_1=T^{2} $ . Here $ S=x^{2}+y^{2}+20(x+y)+20,\ \ S_1=20 $
$ T=10(x+y)+20 $
$ \therefore \ SS_1=T^{2} $
$ \Rightarrow 20,{x^{2}+y^{2}+20(x+y)+20}=10^{2}{{(x+y+2)}^{2}} $
$ \Rightarrow 4x^{2}+4y^{2}+10xy=0\Rightarrow 2x^{2}+2y^{2}+5xy=0 $ .
BETA
