Circle And System Of Circles Question 154
Question: The equation of the circle which passes through the points $ (3,\ -2) $ and $ (-2,\ 0) $ and centre lies on the line $ 2x-y=3 $ , is
[Roorkee 1971]
Options:
A) $ x^{2}+y^{2}-3x-12y+2=0 $
B) $ x^{2}+y^{2}-3x+12y+2=0 $
C) $ x^{2}+y^{2}+3x+12y+2=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let centre be (h, k), then $ {{(h-3)}^{2}}+{{(k+2)}^{2}}={{(h+2)}^{2}}+k^{2} $
$ \Rightarrow 10h-4k-9=0 $ Also the centre lies on the given line, so $ 2h-k=3 $ . On
Solving $ k=-6,\ h=-\frac{3}{2} $ Radius is $ {{(h-3)}^{2}}+{{(k+2)}^{2}}=\frac{145}{4} $ , which is true for option (c) only.
BETA
