How do you write the equation of the circle with center at ( 5, -2) and diameter of 8?

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Answer 1 · 2018-08-03T19:58:40

The eqn. of circle is :

$x^{2} + y^{2} - 10 x + 4 y + 25 = 0$ $x^2+y^2-10x+4y+25=0$

The equation of circle with center at $(h, k)$ $(h,k)$ and radius of $r$ $r$ is :

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $color(red)((x-h)^2+(y-k)^2=r^2$

We have center $(5, - 2) and radius r = \frac{8}{2} = 4$ $(5,-2) and "radius "r=8/2=4$

So ,the eqn. of circle is :

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

$\Rightarrow x^{2} - 10 x + 25 + y^{2} + 4 y + 4 = 4$ $=>x^2-10x+25+y^2+4y+4=4$

$\Rightarrow x^{2} + y^{2} - 10 x + 4 y + 25 = 0$ $=>x^2+y^2-10x+4y+25=0$

Answer 2 · 2018-08-03T23:31:41

${(x - 5)}^{2} + {(y + 2)}^{2} = 16$ $(x-5)^2+(y+2)^2=16$

Recall that the equation of a circle is given by

$bar( ul|color(white)(2/2)(x-h)^2+(y-k)^2=r^2color(white)(2/2)|)$ , with center $(h,k)$ and radius $r$ .

We are centered at $(5,-2)$ , which means $h=5$ and $k=2$ .

We also know that we have a diameter of $8$ , which means our radius is $4$ .

We have all of the information we need, so we can now write the equation of this circle.

$(x-5)^2+(y+2)^2=16$

Hope this helps!