A) \[{{x}^{2}}+{{y}^{2}}-x-y=0\]
B) \[{{x}^{2}}+{{y}^{2}}-6x-4y=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x-2y=0\]
D) \[{{x}^{2}}+{{y}^{2}}+2x+2y=0\]
Correct Answer: C
Solution :
Let the required circle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] ?. (i) This passes through (0, 0), therefore\[c=0\]. The centre \[(-g,\ -f)\] of (i) lies on \[y=x\], hence \[g=f\]. Since (i) cuts the circle \[{{x}^{2}}+{{y}^{2}}-4x-6y+10=0\] orthogonally, therefore \[2(-2g-3f)=c+10\Rightarrow -10g=10\] \[(\because \ g=f\ \text{and}\ c=0)\] \[\Rightarrow g=f=-1\] Hence the required circle is\[{{x}^{2}}+{{y}^{2}}-2x-2y=0\].
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