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: D
Solution :
Given, circle is \[{{x}^{2}}+{{y}^{2}}-2x=0\] ......(i) and line is \[y=x\] .....(ii) Putting \[y=x\] in (i), We get \[2{{x}^{2}}-2x=0\Rightarrow x=0,\,1\] From (i), \[y=0\], 1 Let \[A=(0,\,0)\], \[B=(1,\,1)\] Equation of required circle is \[(x-0)\,(x-1)+(y-0)\,(y-1)=0\] or \[{{x}^{2}}+{{y}^{2}}-x-y=0\].You need to login to perform this action.
You will be redirected in
3 sec