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Question

Find the coordinates of the circumcenter of a triangle whose vertices are (8, 6), (8, – 2) and (2, – 2). What is the circumradius of this triangle?


A
(5,1) and 5
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B
(5,2) and 5
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C
(5,2) and 7
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D
(5,1) and 7
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Solution

The correct option is B (5,2) and 5

Recall that the circumcentre of a traingle is equidistant from the verices of a traingle.
Let A (8 , 6 ), B ( 8, -2 ) and C (2 , -2) be the vertices of the given traingle and let P ( x, y) be the circumcentre of this traingle .
Then, PA = PB = PC
PA2 = PB2 = PC2

Now , PA2 = PB2

(x8)2 + (y6)2 = (x8)2 + (y+2)2

y2 + 36 - 12y = y2 + 4 + 4y

16y = 32

y = 2

and

PB2 = PC2

(x8)2 + (y+2)2 = (x2)2 + (y+2)2

x2 + 64 - 16x = x2 + 4 - 4x

12x = 60

x = 5

So the coordinates of the circumcentre P are (5 ,2)

Also, Circumradius = PA = PB = PC

= ((58)2+(26)2)

= 9+16

= 5 units


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