Circle And System Of Circles Question 79
Question: The equation of a circle with centre $ (-4,\ 3) $ and touching the circle $ x^{2}+y^{2}=1 $ , is
Options:
A) $ x^{2}+y^{2}+8x-6y+9=0 $
B) $ x^{2}+y^{2}+8x+6y-11=0 $
C) $ x^{2}+y^{2}+8x+6y-9=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Centre is $ (-4,\ 3) $ Radius = Distance between centres - Radius of other circle $ =5-1=4 $
Hence equation of circle is $ x^{2}+y^{2}+8x-6y+9=0 $ . Trick: Only option (a) has centre (-4, 3).
BETA
