Circle And System Of Circles Question 163
Question: The locus of the centres of the circles which touch externally the circles $ x^{2}+y^{2}=a^{2} $ and $ x^{2}+y^{2}=4ax $ , will be
Options:
A) $ 12x^{2}-4y^{2}-24ax+9a^{2}=0 $
B) $ 12x^{2}+4y^{2}-24ax+9a^{2}=0 $
C) $ 12x^{2}-4y^{2}+24ax+9a^{2}=0 $
D) $ 12x^{2}+4y^{2}+24ax+9a^{2}=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ C\equiv (h,\ k) $ , radius $ =r $ Co-ordinates of $ A\equiv [ \frac{ah}{a+r},\ \frac{ak}{a+r} ] $
Co-ordinates of $ B\equiv [ \frac{2ar+2ah}{2a+r},\ \frac{2ak}{2a+r} ] $
Putting co-ordinates of A and B in $ S_1,\ S_2 $ respectively and eliminating r, we get the locus $ 12x^{2}-4y^{2}-24ax+9a^{2}=0 $ .
Aliter: Since it touches $ x^{2}+y^{2}=a^{2} $ and $ x^{2}+y^{2}-4ax=0 $ , therefore $ r+a=\sqrt{h^{2}+k^{2}} $ ………….(i) $ r+2a=\sqrt{{{(h-2a)}^{2}}+k^{2}} $ ………….(ii)
From (i), putting the value of r in (ii), we get $ -a+\sqrt{h^{2}+k^{2}}+2a=\sqrt{{{(h-2a)}^{2}}+k^{2}} $
On simplification, we get the required locus.
BETA
