Circle And System Of Circles Question 356
Question: A tangent to the circle $ x^{2}+y^{2}=5 $ at the point (1,-2)….. the circle $ x^{2}+y^{2}-8x+6y+20=0 $
[IIT 1975]
Options:
A) Touches
B) Cuts at real points
C) Cuts at imaginary points
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Tangent is $ x-2y-5=0 $ and points of intersection with circle $ x^{2}+y^{2}-8x+6y+20=0 $ are given by $ 4y^{2}+25+20y+y^{2}-16y-40+6y-20=0 $
$ \Rightarrow 5y^{2}+10y+5=0 $
$ \Rightarrow y=-1 $ and $ x=-3 $ i.e., touches.
BETA
