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) \[{{x}^{2}}+{{y}^{2}}+x-y=0\]
Correct Answer: A
Solution :
The equation of line is\[y=x\] ...(i) and equation of circle is \[{{x}^{2}}+{{y}^{2}}-2x=0\] ...(ii) On solving Eqs. (i) and (ii), we get \[{{x}^{2}}+{{x}^{2}}-2x=0\] \[\Rightarrow \] \[2{{x}^{2}}-2x=0\] \[\Rightarrow \] \[2x(x-1)=0\] \[\Rightarrow \] \[x=0,\text{ }x=1\] When\[x=0,\]then\[y=0\] and when\[x=1,\]then\[y=1\] Let coordinates of A be (0,0)and coordinates of B be (1,1). \[\therefore \]Equation of circle (AB as a diameter) is \[(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})=0\] \[\Rightarrow \] \[(x-0)(x-1)+(y-0)(y-1)=0\] \[\Rightarrow \] \[x(x-1)+y(y-1)=0\] \[\Rightarrow \] \[{{x}^{2}}-x+{{y}^{2}}-y=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-x-y=0\] Alternate Solution Equation of a circle passing through the points of intersection of\[{{x}^{2}}+{{y}^{2}}-2x=0\]and\[y=x\]is \[({{x}^{2}}+{{y}^{2}}-2x)+\lambda (x-y)=0\] i.e., \[{{x}^{2}}+{{y}^{2}}+(\lambda -2)x-\lambda y=0\] Its centre is\[\left( \frac{2-\lambda }{2},\frac{\lambda }{2} \right)\] Since, AB is diameter of required circle. \[\therefore \]\[\left( \frac{2-\lambda }{2},\frac{\lambda }{2} \right)\]must lies on\[y=x\]. \[\Rightarrow \] \[\frac{\lambda }{2}=\frac{2-\lambda }{2}\] \[\Rightarrow \] \[\lambda =1\] \[\therefore \]The equation of required circle is \[{{x}^{2}}+{{y}^{2}}-x-y=0\]
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