find the equation of the circle: y-intercepts 4 and –8, contain (–12, –8)

Answer:[tex]\boxed{(x+6)^2+(y+2)^2=72}[/tex]Step-by-step explanation:The center-radius form of the circle equation is given by:[tex](x-h)^2+(y-k)^2=r^2[/tex]Here we know that the y-intercepts are:[tex]y=4 \ and \ y=-8[/tex]So:[tex]\bullet \ If \ y=4, \ x=0 \\ \\ \\ (0-h)^2+(4-k)^2=r^2 \\ \\ \therefore \mathbf{(I)} \ h^2+16-8k+k^2=r^2 \\ \\ \\ \bullet \ If \ y=-8, \ x=0 \\ \\ \\ (0-h)^2+(-8-k)^2=r^2 \\ \\ \therefore \mathbf{(II)} \ h^2+64+16k+k^2=r^2 \\ \\ \\ \bullet \ If \ x=-12, \ y=-8 \\ \\ \\ (-12-h)^2+(-8-k)^2=r^2 \\ \\ \therefore 144+24h+h^2+64+16k+k^2=r^2 \\ \\ \therefore \mathbf{(III)} \ h^2+208+16k+24h+k^2=r^2[/tex]So we have the following system of equations:[tex]\left\{ \begin{array}{c}(I)\:h^{2}+16-8k+k^{2}=r^{2}\\(II)\:h^{2}+64+16k+k^{2}=r^{2}\\(III)\:h^{2}+208+16k+24h+k^{2}=r^{2}\end{array}\right.[/tex][tex]Subtract \ II \ from \ I \\ \\ \\\left\{ \begin{array}{c}h^{2}+16-8k+k^{2}=r^{2}\\-(h^{2}+64+16k+k^{2}=r^{2})\\---------------------\\h^{2}+16-8k+k^{2}-h^{2}-64-16k-k^{2}=r^{2}-r^{2}\end{array}\right.[/tex][tex]Simplifying: \\ \\ h^{2}+16-8k+k^{2}-h^{2}-64-16k-k^{2}=r^{2}-r^{2}\\ \\ 16-8k-64-16k=0 \\ \\ -24k-48=0 \\ \\ k=-\frac{48}{24} \\ \\ \therefore \boxed{k=-2}[/tex]From (I):[tex]h^2+16-8k+k^2=r^2 \\ \\ \\ For \ k=-2 \\ \\ h^2+16-8(-2)+(-2)^2=r^2 \\ \\ \therefore r^2=h^2+36 \\ \\ \\ Substituting \ k \ and \ r^2 \ into \ (III): \\ \\ h^{2}+208+16(-2)+24h+(-2)^{2}=h^2+36 \\ \\ Simplifying: \\ \\ 180+24h=36 \\ \\ 24h=36-180 \\ \\ 24h=-144 \\ \\ h=-\frac{144}{24} \\ \\ \therefore \boxed{h=-6}[/tex]Finding the radius:[tex]r^2=(-6)^2+36 \\ \\ r^2=36+36 \\ \\ r^2=72 \\ \\ \therefore \boxed{r=6\sqrt{2}}[/tex]Finally, the equation of the circle is:[tex](x-(-6))^2+(y-(-2))^2=72 \\ \\ \boxed{(x+6)^2+(y+2)^2=72}[/tex]