# A triangle has corners at #(4 , 5 )#, #(8 ,2 )#, and #(4 ,7 )#. What is the radius of the triangle's inscribed circle?

##### 1 Answer

#### Explanation:

The first step is to find the location of the center of the circle.

It is where the perpendicular bisectors of the edges of the triangle meet. So it is the intersection of 3 lines that we are finding. But actually 2 lines is enough to find the intersection; the third line **will necessarily** intersect at the same point.

The perpendicular bisector is a locus of points that are equidistant from the corners of the triangle. From the Pythagoras Theorem, an expression for distance from a point with coordinate

#sqrt{(x-p)^2+(y-q)^2}#

So to find the locus of points that are equidistant from

#sqrt{(x-4)^2+(y-5)^2} = sqrt{(x-4)^2+(y-7)^2}#

#(x-4)^2+(y-5)^2 = (x-4)^2+(y-7)^2#

#(y-5)^2 = (y-7)^2#

#y^2-10y+25 = y^2-14y+49#

#4y = 24#

#y = 6#

To find the locus of points that are equidistant from

#sqrt{(x-4)^2+(y-5)^2} = sqrt{(x-8)^2+(y-2)^2}#

#(x-4)^2+(y-5)^2 = (x-8)^2+(y-2)^2#

#(x^2 - 8x + 16) + (y^2 - 10y + 25) = (x^2 - 16x + 64) + (y^2 - 4y + 4)#

# - 8x + 16 - 10y + 25 = - 16x + 64 - 4y + 4#

# 8x = 27 + 6y#

Since we know from the first equation

#x = frac{27 + 6(6)}{8} = 63/8#

The coordinates of the circumcenter is

The radius can be found by calculating the distance from the center of the circle to any of the corners.

#sqrt{(63/8-8)^2 + (6-2)^2} = frac{5sqrt41}{8}#

#sqrt{(63/8-4)^2 + (6-5)^2} = frac{5sqrt41}{8}#

#sqrt{(63/8-4)^2 + (6-7)^2} = frac{5sqrt41}{8}#