The centre of the circle passing through $(6, 0)$ and tangent to the circle $x^{2} + y^{2} = 4$ at $(2, 0)$ is

(0, - 6)

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(1, - 5)

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(0, - 9)

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none

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Solution

The correct option is D none
Two points $(2, 0)$ and $(6, 0)$ as shown in the figure are diametrically opposite.
Solving the question graphically,
the center is $(4, 0)$

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x^{2} + y^{2} = 9

The centre of circle passing through (0, 0) and (1, 0) and touching the circle $x^{2} + y^{2} = 9$ is
(2002)

x^{2} + y^{2} - 2 x + 4 y - 1 = 0

(0, 0)

(2, 0)

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

(0, 3)

S = 0

x^{2} + y^{2} - 2 x + 4 y - 1 = 0

x^{2} + y^{2} + 4 x - 2 y - 5 = 0

x^{2} + y^{2} = 4

S = 0

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