KEAM 2014 Math Paper

Let the vertices of a triangle are A(8,6) B(8,−2) and C(2,−2)Now, ‌‌AB=√(8−8)2+(6+2)2
=√0+82=8
and CA=√(8−2)2+(6+2)2
=√62+82
=√36+64=√100=10
Now, ‌‌AB2+BC2=(8)2+(6)2
=64+36=100=AC2
So, ABC is a right angled triangle and right angled at B.
We know that, in a right angled triangle, the circumcenter is the mid-point of hypotenuse.
∴ Mid-point of AC=(‌

8+2

2

,‌

6−2

2

)=(5,2)
Hence, the required circumcenter is (5,2)