Circle And System Of Circles Question 48
Equation of the tangent to the circle $ x^{2}+y^{2}=a^{2} $ which is perpendicular to the straight line $ y=mx+c $ is $ x = my \pm a\sqrt{m^{2}+1} $
Options:
1
-1
C) $ \frac{3}{2} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
The equation of tangent at point $ (-2,\ -3) $ to the circle $ x^{2}+y^{2}+2x+4y+3=0 $ is, $ -2x-3y+1(x+2)+2(y+3)+3=0 $
$ \Rightarrow -2x-3y+x-2+2y-6+3=0 $
$ \Rightarrow -x-y-5=0\Rightarrow x+y+5=0 $ or $ y=-x-5 $ ; so, $ m=-1 $
Hence, gradient of normal $ =\frac{-1}{m}=1 $ .
BETA
