A) \[{{x}^{2}}+{{y}^{2}}+2{{x}_{1}}(x+y)+x_{1}^{2}=0\]
B) \[{{x}^{2}}+{{y}^{2}}-2{{x}_{1}}(x+y)+x_{1}^{2}=0\]
C) \[{{x}^{2}}+{{y}^{2}}=x_{1}^{2}+y_{1}^{2}\]
D) \[{{x}^{2}}+{{y}^{2}}+2x{{x}_{1}}+2y{{y}_{1}}=0\]
Correct Answer: B
Solution :
The equation will be \[{{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}\] As the circle touches both the axes, \[(r)=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}}=5\] \ \[{{(x-{{x}_{1}})}^{2}}+{{(y-{{x}_{1}})}^{2}}=x_{1}^{2}\] Þ \[{{x}^{2}}+{{y}^{2}}-2{{x}_{1}}(x+y)+x_{1}^{2}=0\].
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