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Equation of a circle which passes through (3, 6) and touches the axes, is

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x2+y2+6x+6y+3=0

x2+y2−6x−6y−9=0

x2+y2−6x−6y+9=0

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detailed solution

Correct option is C

The equation of a circle touching the co-ordinateaxes is(x−a)2+(y−a)2=a2 or, x2+y2−2ax−2ay+a2=0This passes through (3, 6).∴ 9+36−6a−12a+a2=0⇒a=3,15.Hence, the required circle arex2+y2−6x−6y+9=0 or, x2+y2−30x−30y+225=0.

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