Let the vertex A of a triangle ABC be (1, 2), and the mid-point of the side AB be (5,−1). If the centroid of this triangle is (3,4) and its circumcenter is (α,β), then 21(α+β) is equal to:
(A)309
(B)403
(C)497✓ Correct
(D)524
Answer: (C)
Solution
The midpoint of AB gives B=2(5,−1)−(1,2)=(9,−4). From the centroid, C=3(3,4)−A−B=(−1,14). The circumcentre (α,β) is equidistant from the three vertices; equating PA2=PB2 and PA2=PC2 and simplifying yields 4x−3y=23 and 5x−9y=−25. Solving these, α=794 and β=21215. Hence 21(α+β)=21⋅794+21⋅21215=282+215=497.
JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option