Conic Sections Question 228
Question: In the given figure, the equation of the larger circle is $ x^{2}+y^{2}+4y-5=0 $ and the distance between centres is 4. Then the equation of smaller circle is
Options:
A) $ {{(x-\sqrt{7})}^{2}}+{{(y-1)}^{2}}=1 $
B) $ {{(x+\sqrt{7})}^{2}}+{{(y-1)}^{2}}=1 $
C) $ x^{2}+y^{2}=2\sqrt{7}x+2y $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We have $ x^{2}+y^{2}+4y-5=0 $ , Its centre is $ C_1(0,-2), $ $ r_1=\sqrt{4+5}=3. $ Let $ C_2(h,k) $ be the centre of the smaller circle and its radius $ r_2. $ Then $ C_1C_2=4. $
$ \Rightarrow \sqrt{h^{2}+{{(k+2)}^{2}}}=3+r_2=4 $
…… (1)
$ \Rightarrow r_2=1 $ But $ k=r_2=1 $ [it touches x-axis]
$ \therefore $ From eq. (1), $ 4=\sqrt{h^{2}+{{(1+2)}^{2}}} $
$ \Rightarrow 16=h^{2}+9\Rightarrow h^{2}=7\Rightarrow h=\pm \sqrt{7} $ Since $ h>0\therefore h=\sqrt{7} $
Hence, required circle is $ {{(x-\sqrt{7})}^{2}}+{{(y-1)}^{2}}=1 $
BETA
