JEE Main 2025 — 4 April, Morning Shift — Question 43

Solving the line $x+y=3$, and the circle $x^{2}+$ $(y-1)^{2}=2$

Substitute $y=3-x$ : $x^{2}+(3-x-1)^{2}=2$

$\Rightarrow x^{2}-2 x+1=0$

$\Rightarrow x=1 \Rightarrow y=2$

So, $P=\left(x_{1}, y_{1}\right)=(1,2) \Rightarrow x_{1} y_{1}=1 \cdot 2=2$

Use midpoint condition

Let $Q=\left(x_{2}, y_{2}\right), R=\left(x_{3}, y_{3}\right)$.

Since $P$ is the midpoint of $Q R$ :

$x_{2}+x_{3}=2 x_{1}=2, y_{2}+y_{3}=2 y_{1}=4$

So, we can write: $x_{3}=2-x_{2}, y_{3}=4-y_{2}$

Given,

$P Q=\frac{2 \sqrt{2}}{3} \Rightarrow P Q^{2}=\left(x_{2}-1\right)^{2}+\left(y_{2}-2\right)^{2}=\frac{8}{9}$

Let's denote: $x_{2}=a, y_{2}=b, x_{3}=2-a, y_{3}=4-b$

$(a-1)^{2}+(b-2)^{2}=\frac{8}{9}$

$\Rightarrow a^{2}-2 a+1+b^{2}-4 b+4=\frac{8}{9}$

$\Rightarrow a^{2}+b^{2}-2 a-4 b+5=\frac{8}{9}$

$\Rightarrow 9 a^{2}+9 b^{2}-18 a-36 b+37=0$

Hence, $a=\frac{5}{3}, b=\frac{4}{3}$

$x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=2+a b+(2-a)(4-b)$

$9\left(x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}\right)=9(10+2 a b-2 b-4 a)$

$=90+18 a b-18 b-36 a=46$