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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.

Oct 23, 2019

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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

Oct 23, 2019