# The centre of circle passing through the points (0,0) and (1,0) and touching the circle x^2+y^2=9 is

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by bkritagya

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by bkritagya

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x² + y² + 2gx + 2fy + c = 0

centre = (-g,-f)

radius = √(g²+f²-c)

it passes through (0,0)&(1,0) so point satisfy the equation

put (0,0)

c = 0

put (1,0)

g = -1/2

since the circle touches internally hence

distance between centre = difference of radius

centre of given circle = (0,0)

radius = 3

√(g² + f²) = 3 - √(g²+f²-c)

√(1/4 + f²) = 3 - √(1/4 + f²)

√(1/4 + f²) = 3/2

1/4 + f² = 9/4

f² = 2

f = √2 or -√2

hence centre = (1/2,√2) or (1/2,-√2)