What is the center and radius of the circle with equation $36 x^{2} - 36 x + 36 y^{2} - 24 y = 131$ $36x^2 - 36x + 36y^2 - 24y = 131$ ?

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What is the center and radius of the circle with equation $36 x^{2} - 36 x + 36 y^{2} - 24 y = 131$ $36x^2 - 36x + 36y^2 - 24y = 131$ ?

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Answer 1 · 2016-01-29T17:34:41

The General equation of circle is : $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ $x^2 + y^2 + 2gx + 2fy + c = 0$

Explanation:

So,
Dividing all terms by 36 ( To remove the coefficient of x^2 and y^2)
We get,
$x^{2} - x + y^{2} + \frac{2}{3} y = \frac{131}{36}$ $x^2 - x + y^2 + 2/3 y = 131/36$
Now
$x^{2} - x + y^{2} + \frac{2}{3} y - \frac{131}{36} = 0$ $x^2 - x + y^2 + 2/3 y- 131/36 = 0$

Comparing the given equation with the general equation of circle we get,

2g = $- 1$ $-1$
or g = $- \frac{1}{2}$ $-1/2$

Again,
2f = $\frac{2}{3}$ $2/3$
or f = $\frac{1}{3}$ $1/3$

And,

c= $- \frac{131}{36}$ $-131/36$

Thus,
Center of the circle (h,k) = ( $- g, - f$ $-g ,-f$ )
= ( $\frac{1}{2}, - \frac{1}{3}$ $1/2 , -1/3$ )
And
Radius = $\sqrt{g^{2} + f^{2} - c}$ $sqrt(g^2+f^2-c)$

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What is the center and radius of the circle with equation $36 x^{2} - 36 x + 36 y^{2} - 24 y = 131$ $36x^2 - 36x + 36y^2 - 24y = 131$ ?

1 Answer

Explanation:

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What is the center and radius of the circle with equation 36x^2 - 36x + 36y^2 - 24y = 13136x2−36x+36y2−24y=13136x^2 - 36x + 36y^2 - 24y = 131?

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Explanation:

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What is the center and radius of the circle with equation $36 x^{2} - 36 x + 36 y^{2} - 24 y = 131$ $36x^2 - 36x + 36y^2 - 24y = 131$ ?