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Equation of a circle which passes through (3, 6) and touches the axes is
A. x2 + y+ 6x + 6y + 3 = 0
B. x2 + y2 – 6x – 6y – 9 = 0
C. x+ y2 – 6x – 6y + 9 = 0
D. none of these

Answers (1)

Given that the circle touches both axes

Therefore, equation of the circle is (x-a)2+(y-a)2=a2

circle passes through (3,6)

(3-a)2+(6-a)2=a2

a2-18a+45=0

(a-3)(a-15)=0

a = 3 or a =15

For a=3 equation of circle

(x-3)2+(y-3)2=9

x2+y2-6x-6y+9=0

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