Question

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

The radius of the circle is
units. The point (-15,
) lies on this circle

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 = √(( x_2- x_1)^2 + ( y_2- y_1)^2)\\\\\\\stackrel{radius}{r}=√([8-(-7)]^2+[7-(-1)]^2)\implies r=√((8+7)^2+(7+1)^2)\\\\\\r=√(15^2+8^2)\implies r=√(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.

Answer 2

The radius of the circle is 17 units. The point (-15, 14 ) lies on this circle.

Explanation:

The onther guys was right ;)