The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is

(x - 5)^{2} + (y + 5)^{2} = 25

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(x + 4)^{2} + (y + 4)^{2} = 25

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(x - 6)^{2} + (y + 6)^{2} = 25

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(x + 5)^{2} + (y + 5)^{2} = 25

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Solution

The correct option is D $(x + 5)^{2} + (y + 5)^{2} = 25$
Since circle touches the co-ordinate axes in III quadrant.

$∴$ Radius = -h = -k. Hence h = k = -5
$∴$ Equation of circle is $(x + 5)^{2} + (y + 5)^{2} = 25.$

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x = 5 - 5 sin t

y = 4 + 5 cos t

25 {{(x - 2)}^{2} + {(y + 5)}^{2}} = {(3 x + 4 y - 1)}^{2}

\frac{25}{k} = 1^{2} - \frac{2^{2}}{5} + \frac{3^{2}}{5^{2}} - \frac{4^{2}}{5^{3}} + \frac{5^{2}}{5^{4}} - \frac{6^{2}}{5^{2}} + . . . . . . \infty

k

3 x + 4 y - 5 = 0

3 x + 4 y + 25 = 0

x + 2 y = 0

{(x - 4)}^{2} + {(y - 5)}^{2} = 25

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