JEE Main 2025 — 22 January, Evening Shift — Question 22

All the three points A, B, C lie on the circle $x^{2}+y^{2}=100$ so circumcentre is $(0,0)$

$\underset{\mathrm{O}(0,0)}{\stackrel{1}{\mathrm{G}(\mathrm{h}, \mathrm{k})} \quad \mathrm{L}(\mathrm{a}, 9)}$

$\frac{\mathrm{a}+0}{3}=\mathrm{h} \Rightarrow \mathrm{a}=3 \mathrm{~h}$

and $\frac{9+0}{3}=k \Rightarrow k=3$

also centroid $\frac{6+10 \cos \alpha-10 \sin \alpha}{3}=\mathrm{h}$

$\Rightarrow 10(\cos \alpha-\sin \alpha)=3 \mathrm{~h}-6..(i)$

and $\frac{8+10 \cos \alpha-10 \sin \alpha}{3}=\mathrm{k}$

$\Rightarrow 10(\cos \alpha-\sin \alpha)=3 \mathrm{k}-8=9-8=1…(ii)$ on squaring $100(1-\sin 2 \alpha)=1$

$\Rightarrow 100 \sin 2 \alpha=99$

from equ. (i) and (ii) we get $\mathrm{h}=\frac{7}{3}$

Now $5 \mathrm{a}-3 \mathrm{~h}+6 \mathrm{k}+100 \sin 2 \alpha$

$=15 \mathrm{~h}-3 \mathrm{~h}+6 \mathrm{k}+100 \sin 2 \alpha$

$=12 \times \frac{7}{3}+18+99$

$=145$