Question

A circle is centered at the point (-7, -1) and passes through the point (8, 7).

Answer 1

we know the circle's center is at -7, -1, and we know the circle itself passes through 8,7, the distance from the center to a point on it is by definition its radius, therefore

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\\stackrel{center}{(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-1})}\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\\stackrel{radius}{r}=\sqrt{[8-(-7)]^2+[7-(-1)]^2}\implies r=\sqrt{(8+7)^2+(7+1)^2}\\\\\\r=\sqrt{15^2+8^2}\implies r=\sqrt{225+64}\implies \boxed{r=17}$

and since we know that x = -15 for such a point, then

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2\qquad center~~(\stackrel{-7}{ h},\stackrel{-1}{ k})\qquad \qquad radius=\stackrel{17}{ r}\\\\\\\[x-(-7)]^2+[y-(-1)]^2=17^2\implies (x+7)^2+(y+1)^2=289\\\\\\\stackrel{\textit{since we know x = -15}}{(-15+7)^2+(y+1)^2=289}\implies (-8)^2+\stackrel{FOIL}{(y^2+2y+1^2)}=289\\\\\\64+y^2+2y+1=289\implies y^2+2y=224\implies y^2+2y-224=0\\\\\\(y+16)(y-14)=0\implies y=\begin{cases}-16\\14\end{cases}$

since it's a circle, it touches x = -15 twice, check the picture below.