The equation of a circle passing through ( 3, - 6) and touching both the axes, is
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a
x2+y2−6x+6y+9=0
b
x2+y2+6x−6y+9=0
c
x2+y2+30x−30y+225=0
d
x2+y2+30x+30y+225=0
answer is A.
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Detailed Solution
Circles given in option (a) passes through ( 3, - 6) and touches both the axes. ALITER The circle touching both the axes in the fourth quadrant is x2+y2−2ax+2ay+a2=0,a>0.It passes through (3,-6). ∴ 45−6a−12a+a2=0⇒a2−18a+45=0⇒a=3,5So, the equations are x2+y2−6x+6y+9=0 and x2+y2−10x+10y+25=0