+817 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.
+9481 I'm sure there is an easier way to do this, but here is one way to get the answer....
The intersection point of y = -x - 1 and x = 2 is (2, -3)
The intersection point of y = -x - 1 and y = 1/5x + 13/5 is (-3, 2)
The intersection point of x = 2 and y = 1/5x + 13/5 is (2, 3)
equation of a circle: (x - h)2 + (y - k)2 = r2
Using the point (2, -3) we can make the equation: (2 - h)2 + (-3 - k)2 = r2
Using the point (-3, 2) we can make the equation: (-3 - h)2 + (2 - k)2 = r2
Using the point (2, 3) we can make the equation: (2 - h)2 + (3 - k)2 = r2
Set the first and third values of r2 equal to each other.
(2 - h)2 + (-3 - k)2 = (2 - h)2 + (3 - k)2
(-3 - k)2 = (3 - k)2
-3 - k = ±(3 - k)
-3 - k = 3 - k or -3 - k = -(3 - k)
not a solution -3 - k = -3 + k
k = 0
Use this value of k and set the first and second values of r2 equal to each other.
(2 - h)2 + (-3 - 0)2 = (-3 - h)2 + (2 - 0)2
(2 - h)2 + 9 = (-3 - h)2 + 4
(2 - h)2 + 5 = (-3 - h)2
(2 - h)2 - (-3 - h)2 = -5
(4 - 4h + h2) - (9 - 6h + h2) = -5
-5 + 2h = -5
2h = 0
h = 0
Now we can find r2 .
(2 - 0)2 + (-3 - 0)2 = r2
13 = r2
So the equation of the circle is...
x2 + y2 = 13
Here's a graph to check this: https://www.desmos.com/calculator/tcsq7scrno
