A) \[{{x}^{2}}+{{y}^{2}}+5x+5y+25=0\]
B) \[{{x}^{2}}+{{y}^{2}}-10x-10y+25=0\]
C) \[{{x}^{2}}+{{y}^{2}}-5x-5y+25=0\]
D) \[{{x}^{2}}+{{y}^{2}}+10x+10y+25=0\]
Correct Answer: B
Solution :
The centre of the circle which touches each axis in first quadrant at a distance 5, will be (5, 5) and radius will be 5. \[\therefore \]\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{a}^{2}}\]\[\Rightarrow {{(x-5)}^{2}}+{{(y-5)}^{2}}={{(5)}^{2}}\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}-10x-10y+25=0\].
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