Straight LinesmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Circumcenter of a Triangle: 21(α+β) = 497 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let the vertex AA of a triangle ABCABC be (1,2)(1,2), and the mid-point of the side ABAB be (5,−1)(5,-1). If the centroid of the triangle is (3,4)(3,4) and its circumcenter is (α,β)(\alpha,\beta), then 21(α+β)21(\alpha+\beta) is equal to
A309309
B403403
C497497correct
D524524
Solution
Step 1: Midpoint of AB=(5,−1)AB=(5,-1) with A(1,2)A(1,2) gives B=(9,−4)B=(9,-4). Centroid (3,4)(3,4) gives CC: 1+9+Cx3=3⇒Cx=−1\dfrac{1+9+C_x}{3}=3\Rightarrow C_x=-1; 2−4+Cy3=4⇒Cy=14\dfrac{2-4+C_y}{3}=4\Rightarrow C_y=14. So C(−1,14)C(-1,14). Step 2: Circumcenter P(α,β)P(\alpha,\beta) is found from perpendicular bisectors. Their equations come out as
4x−3y=23and5x−9y=−25.4x-3y=23\qquad\text{and}\qquad 5x-9y=-25.
 Step 3: Solve: P=(947, 21521)P=\left(\dfrac{94}{7},\ \dfrac{215}{21}\right). Step 4:
21(α+β)=21(947+21521)=3⋅94+215=282+215=497.21(\alpha+\beta)=21\left(\frac{94}{7}+\frac{215}{21}\right)=3\cdot94+215=282+215=497.
 Correct answer: (3)
Solution working
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