A) \[{{x}^{2}}-{{y}^{2}}-x-y=0\]
B) \[{{x}^{2}}+{{y}^{2}}-x-y=0\]
C) \[{{x}^{2}}+{{y}^{2}}+x+y=0\]
D) \[{{x}^{2}}+{{y}^{2}}-2x-2y=0\]
Correct Answer: B
Solution :
Let the equation of the required circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] ... (i) This passes through\[A(1,\,\,0)\]and\[B(0,\,\,1)\], therefore, \[1+2g+c=0\]and\[1+2f+c=0\] \[\Rightarrow \] \[g=-\left( \frac{c+1}{2} \right)\]and\[f=-\left( \frac{c+1}{2} \right)\] Let\[r\]be the radius of circle (i), then \[r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\] \[\Rightarrow \] \[r=\sqrt{{{\left( \frac{c+1}{2} \right)}^{2}}+{{\left( \frac{c+1}{2} \right)}^{2}}-c}\] \[\Rightarrow \] \[r=\sqrt{\frac{{{c}^{2}}+1}{2}}\] \[\Rightarrow \] \[{{r}^{2}}=\frac{1}{2}=({{c}^{2}}+1)\] Clearly,\[r\]is minimum when\[c=0\]and the minimum value of\[r\]is\[\frac{1}{\sqrt{2}}\]. For\[c=0\], we have \[g=-\frac{1}{2}\]and\[f=-\frac{1}{2}\] On substituting the values of\[g,\,\,f\]and\[c\]in Eq. (i), we get, \[{{x}^{2}}+{{y}^{2}}-x-y=0\] This is equation of the required circle.
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