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) \[{{x}^{2}}+{{y}^{2}}-2x-4y+8=0\]
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 \[{{(x-1)}^{2}}+{{(y-2)}^{2}}={{(\sqrt{2})}^{2}}\] \[\Rightarrow \,{{x}^{2}}+{{y}^{2}}-2x-4y+3=0.\]You need to login to perform this action.
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