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# Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, – 2) and (2, – 2). Also, find its circumradius. [3 MARKS]

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Solution

## Concept : 1 Mark Application : 1 Mark Calculation : 1 Mark Recall that the circumcentre of a traingle is equidistatnt 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 ⇒(x−8)2+(y−6)2=(x−8)2+(y+2)2 ⇒x2+y2−16x−12y+100=x2+y2−16x+4y+68 ⇒16y=32 ⇒y=2 and PB2=PC2 ⇒(x−8)2+(y+2)2=(x−2)2+(y+2)2 ⇒x2+y2−16x+4y+68=x2+y2−4x+4y+8 ⇒12x=60 ⇒x=5 So the coordinates of the circumcentre P are (5 ,2) Also, Circum- radius = PA = PB = PC =√((5−8)2+(2−6)2) =√9+16 =5

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