Circle And System Of Circles Question 201
Question: The equations of tangents to the circle $ x^{2}+y^{2}-22x-4y+25=0 $ which are perpendicular to the line $ 5x+12y+8=0 $ are
Options:
A) $ 12x-5y+8=0 $ , $ 12x-5y=252 $
B) $ 12x-5y=0,12x-5y=252 $
C) $ 12x-5y-8=0,,12x-5y+252=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of line perpendicular to $ 5x+12y+8=0 $ is $ 12x-5y+k=0 $ .
Now it is a tangent to the circle, if
Radius of circle = Distance of line from centre of circle $ \sqrt{121+4-25}=| \frac{12(11)-5(2)+k}{\sqrt{144+25}} | $
$ \Rightarrow k=8 $ or - 252.
Hence equations of tangents are $ 12x-5y+8=0 $ and $ 12x-5y=252 $ .
BETA
