How do you write the equation of the circle with diameter that has endpoints at (–1, –6) and (–3, –2).?

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How do you write the equation of the circle with diameter that has endpoints at (–1, –6) and (–3, –2).?

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Answer 1 · 2016-06-30T15:44:06

$x^{2} + y^{2} + 4 x + 8 y = 0$ $x^2+y^2+4x+8y=0$

Explanation:

The center of the circle is the middle point of the diameter.

Then its coordinates are obtained by using

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ $((x_1+x_2)/2,(y_1+y_2)/2)$

in this case:

$(\frac{- 1 - 3}{2}; \frac{- 6 - 2}{2})$ $((-1-3)/2;(-6-2)/2)$

$(- 2; - 4)$ $(-2;-4)$

The radius is the half of the diameter, that is obtained by calculating the distance

$\sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

in this case:

$\sqrt{{(- 3 + 1)}^{2} + {(- 2 + 6)}^{2}}$ $sqrt((-3+1)^2+(-2+6)^2)$

$\sqrt{4 + 16} = \sqrt{20}$ $sqrt(4+16)=sqrt(20)$

To obtain teh equation of the circle you can substitute in the general equation $x_{0} = - 2, y_{0} = - 4, r = \sqrt{20}$ $x_0=-2, y_0=-4, r=sqrt(20)$ :

(the general equation is ${(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} = r^{2}$ $(x-x_0)^2+(y-y_0)^2=r^2$ )

${(x + 2)}^{2} + {(y + 4)}^{2} = 20$ $(x+2)^2+(y+4)^2=20$

$x^{2} + 4 + 4 x + y^{2} + 16 + 8 y - 20 = 0$ $x^2+4+4x+y^2+16+8y-20=0$

$x^{2} + y^{2} + 4 x + 8 y = 0$ $x^2+y^2+4x+8y=0$

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How do you write the equation of the circle with diameter that has endpoints at (–1, –6) and (–3, –2).?

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