Question

Equation of the circle which of the mirror image of the circle $ {{x}^{2}}+{{y}^{2}}-2x=0 $ in the line $ x+y=2 $ is

• A. $ {{x}^{2}}+{{y}^{2}}-2x+4y+3=0 $
• B. $ 2({{x}^{2}}+{{y}^{2}})+x+y+1=0 $
• C. $ {{x}^{2}}+{{y}^{2}}-4x-2y+4=0 $
• D. None of the above

Answer 1

Answer: (C)

Given equation of circle and line are $ {{x}^{2}}+{{y}^{2}}-2x=0 $ .. (i) and $ x+y=2 $ ..(ii) Centre and radius of the circle are $ (1,0) $ and1. Let the centre of the image circle be $ ({{x}_{1}}{{y}_{1}}) $ . Hence, $ ({{x}_{1}}{{y}_{1}}) $ be the image of the point $ (1,0) $ with respect to the line

$=x+y=2, $ then $ \frac{{{x}_{1}}-1}{1}=\frac{{{y}_{1}}-0}{1}=\frac{-2[1(1)+1(0)-2]}{{{(1)}^{2}}+{{(1)}^{2}}} $

$ \Rightarrow $ $ \frac{{{x}_{1}}-1}{1}=\frac{{{y}_{1}}}{1}=\frac{2}{2}=1 $

$ \Rightarrow $ $ {{x}_{1}}=2,\,\,{{y}_{1}}=1 $

$ \therefore $ Equation of the imaged circle is $ {{(x-2)}^{2}}+{{(y-1)}^{2}}={{1}^{2}} $

$ \Rightarrow $ $ {{x}^{2}}+4-4x+{{y}^{2}}+1-2y=1 $

$ \Rightarrow $ $ {{x}^{2}}+{{y}^{2}}-4x-2y+4=0 $