Circle And System Of Circles Question 196
Question: The equation of circle with centre (1, 2) and tangent $ x+y-5=0 $ is
[MP PET 2001]
Options:
A) $ x^{2}+y^{2}+2x-4y+6=0 $
B) $ x^{2}+y^{2}-2x-4y+3=0 $
C) $ x^{2}+y^{2}-2x+4y+8=0 $
D) $ x^{2}+y^{2}-2x-4y+8=0 $
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Answer:
Correct Answer: B
Solution:
$ \because $ Radius of circle = perpendicular distance of tangent from the centre of circle
$ \Rightarrow $ $ r=\frac{1+2-5}{\sqrt{1+1}}=\sqrt{2} $
Hence the equation of required circle is $ {{(x-1)}^{2}}+{{(y-2)}^{2}}={{(\sqrt{2})}^{2}} $
$ \Rightarrow ,x^{2}+y^{2}-2x-4y+3=0. $
BETA
