Exercise 27 Page 580

We are given the vertices of $△XYZ$ and asked to evaluate the equation of the circle circumscribed about this triangle. First, let's draw this triangle in the coordinate plane.

Now, let's recall that the center of the circle will be the point of intersection of the perpendicular bisectors of the sides of the inscribed triangle. First we can find the perpendicular bisector of side $XY.$ To do this we should evaluate the midpoint of this side. The midpoint of side $XY$ occurs at point $(4,9).$ Since this side is vertical the line perpendicular to this side will be horizontal. Therefore the perpendicular bisector of this side is the line $y=9.$ Next we will find the perpendicular bisector of side $YZ.$ Again we will start with evaluating the midpoint of this side. The midpoint of side $YZ$ occurs at point $(6,11).$ Next we will evaluate the slope of this side. To do this we can use the Slope Formula and substitute points $Y$ and $Z.$ The slope of side $YZ$ is $\text{-}1.$ Now, let's recall that the product of the slopes of the perpendicular segments is always equal to $\text{-}1.$ Using this information, we can conclude that the slope of the perpendicular bisector of this side is $1.$ Using this information, we can write a partial equation for the line that is a perpendicular bisector of side $YZ.$ By substituting the midpoint $(6,11)$ we can find the value of $b.$ The equation of line is $y=x+5.$ As we can see, the perpendicular bisectors of the sides of this triangle intersect at point $(4,9),$ and this point is the center of the circle circumscribed about $△XYZ.$ To evaluate the radius we will evaluate the distance between the center $(4,9)$ and any vertex. We can choose $X(4,5).$ The radius of the circle is $4.$ Finally, we can write the equation of this circle by substituting the center point and the radius into the standard equation of a circle.