Circle And System Of Circles Question 6
Question: The equation of pair of tangents to the circle $ x^{2}+y^{2}-2x+4y+3=0 $ from $ (6,-5) $ , is
[AMU 1980]
Options:
A) $ 7x^{2}+23y^{2}+30xy+66x+50y-73=0 $
B) $ 7x^{2}+23y^{2}+30xy-66x-50y-73=0 $
C) $ 7x^{2}+23y^{2}-30xy-66x-50y+73=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ SS_1=T^{2} $
$ \Rightarrow (x^{2}+y^{2}-2x+4y+3)(36+25-12x-20y+3) $
$ ={{(6x-5y-x-6+2(y-5)+3)}^{2}} $
$ \Rightarrow 7x^{2}+23y^{2}+30xy+66x+50y-73=0 $ .
BETA
