Equation of the circle of the radius 5, and touching the co-ordinate axes in the third quadrant is

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

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(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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Solution

The correct option is B $(x + 5)^{2} + (y + 5)^{2} = 25$
As Given, circles touching the coordinate axes in the $3^{r d}$ Quadrant and having radius $5$ . Therefore from the figure we can see that the coordinate of the center is $(- 5, - 5)$
then equation of that circle is,
$(x - h)^{2} + (y - k)^{2} = r^{2}$ where $(h, k)$ are co-ordinate of the center of the circle
$\Rightarrow (x + 5)^{2} + (y + 5)^{2} = 25$

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Similar questions

x = 5 - 5 sin t

y = 4 + 5 cos t

3 x + 4 y - 5 = 0

3 x + 4 y + 25 = 0

x + 2 y = 0

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

(h, k)

x -

3

8

y -

(h + k)

\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

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