Circle And System Of Circles Question 273
Question: The locus of the centre of a circle which cuts orthogonally the circle $ x^{2}+y^{2}-20x+4=0 $ and which touches $ x=2 $ is
[UPSEAT 2001]
Options:
A) $ y^{2}=16x+4 $
B) $ x^{2}=16y $
C) $ x^{2}=16y+4 $
D) $ y^{2}=16x $
Show Answer
Answer:
Correct Answer: D
Solution:
Let the circle be $ x^{2}+y^{2}+2gx+2fy+c=0 $ …..(i) It cuts the circle $ x^{2}+y^{2}-20x+4=0 $ orthogonally
$ \therefore $ $ 2(-10g+0\times f)=c+4 $
$ \Rightarrow $ $ -20g,=,c+4 $ …..(ii) Circle (i) touches the line $ x=2 $ , \ $ x+0y-2=0 $
$ \therefore $ $ | \frac{-g+0-2}{\sqrt{1^{2}+0^{2}}} |=\sqrt{g^{2}+f^{2}-c} $
$ \Rightarrow $ $ {{(g+2)}^{2}}=g^{2}+f^{2}-c $
$ \Rightarrow $ $ 4g+4=f^{2}-c $ …..(iii)
Eliminating c from (ii) and (iii), we get $ -16g-4=f^{2}-4 $
$ \Rightarrow $ $ f^{2}+16g=0 $
Hence the locus of $ (-g,,-f) $ is $ y^{2}-16x=0 $ .
BETA
