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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.
 

 Mar 6, 2018
 #1
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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

 Mar 6, 2018

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