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
Correct Answer: C
Solution :
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 \[{{(x-2)}^{2}}+{{(y-1)}^{2}}={{1}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+4-4x+{{y}^{2}}+1-2y=1\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4x-2y+4=0\]
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