Circle And System Of Circles Question 143
Question: The equation of the circle with centre at (1, -2) and passing through the centre of the given circle $ x^{2}+y^{2}+2y-3=0 $ , is
Options:
A) $ x^{2}+y^{2}-2x+4y+3=0 $
B) $ x^{2}+y^{2}-2x+4y-3=0 $
C) $ x^{2}+y^{2}+2x-4y-3=0 $
D) $ x^{2}+y^{2}+2x-4y+3=0 $
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Answer:
Correct Answer: A
Solution:
According to the question, the required circle passes through (0,-1).
Therefore, the radius is the distance between the points (0, -1) and (1, -2) i.e., $ \sqrt{2} $ .
Hence the equation is $ {{(x-1)}^{2}}+{{(y+2)}^{2}}={{(\sqrt{2})}^{2}} $
$ \Rightarrow x^{2}+y^{2}-2x+4y+3=0 $
Trick : Since this is the only circle passing through (0, -1).
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