A circle of radius $2$ lies in the first quadrant and touches both the axes of co-ordinates. Then the equation of the circle with centre $(6, 5)$ and touching the above circle externally is

(x - 6)^{2} + (y - 5)^{2} = 4

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(x - 6)^{2} + (y - 5)^{2} = 9

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(x - 6)^{2} + (y - 5)^{2} = 36

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none of these

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Solution

The correct option is D $(x - 6)^{2} + (y - 5)^{2} = 4$
If $(h, k)$ is the center and the radius is $r$ then the equation of the circle is given by $(x - h)^{2} + (y - k)^{2} = r^{2}$

Given that The center of the circle $(h, k) = (6, 5)$ and the radius $r = 2$

Therefore, the equation of the circle is $(x - 6)^{2} + (y - 5)^{2} = 4$

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\frac{x}{a} + \frac{y}{b} = 1

(x - 2)^{2} + (y - 5)^{2} = a^{2}

(x - 3)^{2} + (y - 6)^{2} = b^{2}

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