JEE Main 2026 — 4 April, Morning Shift — Question 35

Let $A(1,2)$, midpoint of AB is $M(5,-1)$. Then $B = 2M - A = (10-1, -2-2) = (9,-4)$. Let centroid $G(3,4)$. For triangle ABC, $G = \left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}\right)$. Thus $\frac{1+9+x_C}{3}=3 \Rightarrow x_C=-1$, and $\frac{2-4+y_C}{3}=4 \Rightarrow y_C=14$. So $C(-1,14)$. Now find circumcenter $O(\alpha,\beta)$ of triangle ABC. It is the intersection of perpendicular bisectors of any two sides. Midpoint of AB is $M(5,-1)$, slope of AB = $\frac{-4-2}{9-1} = \frac{-6}{8} = -\frac{3}{4}$.

So slope of perpendicular bisector of AB = $\frac{4}{3}$.

Equation: $y+1 = \frac{4}{3}(x-5) \Rightarrow 3y+3 = 4x-20 \Rightarrow 4x-3y=23$. Midpoint of BC: $B(9,-4), C(-1,14)$ gives midpoint $N(4,5)$.

Slope of BC = $\frac{14+4}{-1-9} = \frac{18}{-10} = -\frac{9}{5}$.

So slope of perpendicular bisector of BC = $\frac{5}{9}$. Equation: $y-5 = \frac{5}{9}(x-4) \Rightarrow 9y-45 = 5x-20 \Rightarrow 5x-9y = -25$. Solve $4x-3y=23$ and $5x-9y=-25$. Multiply first by 3: $12x-9y=69$.

Subtract second: $(12x-9y)-(5x-9y)=69-(-25) \Rightarrow 7x=94 \Rightarrow x = \frac{94}{7}$.

Then $4\cdot\frac{94}{7} - 3y = 23 \Rightarrow \frac{376}{7} - 3y = 23 \Rightarrow 3y = \frac{376}{7} - \frac{161}{7} = \frac{215}{7} \Rightarrow y = \frac{215}{21}$.

So $\alpha = \frac{94}{7}, \beta = \frac{215}{21}$. Then $\alpha+\beta = \frac{94}{7} + \frac{215}{21} = \frac{282}{21} + \frac{215}{21} = \frac{497}{21}$.

Hence $21(\alpha+\beta) = 497$.