How do you write an equation of a circle with center at (-3,-10), d=24?

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Answer 1 · 2018-06-02T21:03:55

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

Where $(a, b)$ $(a,b)$ are the coordinates of the center.

d=2r

24= 2r

$\frac{24}{2} = r$ $24/2=r$

r=12

Substituting values,

${(x + 3)}^{2} + {(y + 10)}^{2} = 144$ $(x+3)^2+(y+10)^2=144$

$x^{2} + 6 x + 9 + y^{2} + 20 y + 100 = 144$ $x^2+6x+9+y^2+20y+100=144$

$x^{2} + y^{2} + 6 x + 20 y + 100 + 9 - 144 = 0$ $x^2+y^2+6x+20y+100+9-144=0$

$x^{2} + y^{2} + 6 x + 20 y - 35 = 0$ $x^2+y^2+6x+20y-35=0$