How do you find the center and radius of the circle given by the equation $x^{2} + y^{2} - 8 x - 6 y + 21 = 0$ $x^2+y^2-8 x- 6 y +21=0$ ?

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How do you find the center and radius of the circle given by the equation $x^{2} + y^{2} - 8 x - 6 y + 21 = 0$ $x^2+y^2-8 x- 6 y +21=0$ ?

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Answer 1 · 2016-03-02T20:32:30

centre = (4 , 3 ) , r = 2

Explanation:

The general equation of a circle is :

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ $x^2 + y^2 + 2gx + 2fy + c = 0$

centre = ( -g , -f ) and r = $\sqrt{g^{2} + f^{2} - c}$ $sqrt(g^2+f^2 -c)$

compare this with $x^{2} + y^{2} - 8 x - 6 y + 21 = 0$ $x^2 + y^2 - 8x -6y + 21 = 0$

hence: 2g = -8 → g=-4 , 2f =-6 → f=-3 and c=21

centre=(4,3) and r = $\sqrt{{(- 4)}^{2} + {(- 3)}^{2} - 21} = \sqrt{4} = 2$ $sqrt((-4)^2+(-3)^2-21) = sqrt4 = 2$

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How do you find the center and radius of the circle given by the equation $x^{2} + y^{2} - 8 x - 6 y + 21 = 0$ $x^2+y^2-8 x- 6 y +21=0$ ?

1 Answer

Explanation:

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How do you find the center and radius of the circle given by the equation x2+y2−8x−6y+21=0x^2+y^2-8 x- 6 y +21=0?

1 Answer

Explanation:

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How do you find the center and radius of the circle given by the equation $x^{2} + y^{2} - 8 x - 6 y + 21 = 0$ $x^2+y^2-8 x- 6 y +21=0$ ?