The vertices of a triangle are the points of intersection of the line $y = -x-1$, the line $x=2$, and $y = \frac{1}{5}x+\frac{13}{5}$. Find an equation of the circle passing through all three vertices.

Guest Oct 23, 2019

#1**+1 **

The vertices of a triangle are the points of intersection of the line y = -x-1, the line x=2, and y = (1/5)x + 13/5. Find an equation of the circle passing through all three vertices.

We are looking for the equation of the circumcircle of the triangle

The vertices of the triangle are ( -3,2) (2,3) and ( 2,-3)

We need to find the circumcenter thusly :

The midpoint of (2,3) and (2,-3) is [ (2+2)/2, (3 -3)/2] = (2, 0)

The equation of the perpendcular bisector of the side of the triangle joining( (2,3) and (2,-3) is

y = 0 (1)

The midpont of (-3, 2) and ( 2.-3) is [ (-3 + 2)/2 , (-3 + 2)/2 ] = (-1/2, -1/2)

The slope of the line joining (-3,2) and (2,-3) is (-3 - 2) /(2 - -3) = (-5)/(5) = -1

So.......the perpendicular bisector of the side of the triangle containing (-3,2) and (2.-3) has the equation

y = (x +1/2 ) -1/2

y = x (2)

The intersection of (1) and (2) is (0,0)

This is the center of the circumcircle

The radius of the circumcircle is sqrt [ 2^2 + 3^2 ] = sqrt (13)

So.......the equation of the circumcircle is

x^2 + y^2 = 13

See the graph here : https://www.desmos.com/calculator/q80p7ep7im

CPhill Oct 23, 2019