Circle And System Of Circles Question 166
Question: The locus of the centre of a circle of radius 2 which rolls on the outside of circle $ x^{2}+y^{2}+3x-6y-9=0 $ , is
Options:
A) $ x^{2}+y^{2}+3x-6y+5=0 $
B) $ x^{2}+y^{2}+3x-6y-31=0 $
C) $ x^{2}+y^{2}+3x-6y+\frac{29}{4}=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let (h, k) be the centre of the circle which rolls on the outside of the given circle.
Centre of the given circle is $ ( \frac{-3}{2},\ 3 ) $ and its radius $ =\sqrt{\frac{9}{4}+9+9}=\frac{9}{2} $ .
Clearly, (h, k) is always at a distance equal to the sum $ ( \frac{9}{2}+2 ) $
$ =\frac{13}{2} $ of the radii of two circles from $ ( -\frac{3}{2},\ 3 ) $ . Therefore $ {{( h+\frac{3}{2} )}^{2}}+{{(k-3)}^{2}}={{( \frac{13}{2} )}^{2}} $
$ \Rightarrow h^{2}+k^{2}+3h-6k+\frac{9}{4}+9-\frac{169}{4}=0 $
$ \Rightarrow $ Hence locus of (h, k) is $ x^{2}+y^{2}+3x-6y-31=0 $ .
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