JEE Main 2025 — Circle Question with Solution

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

Substitute $y=3-x$ :

$$\begin{array}{rl}& x^2+(3-x-1{)}^2=2\\ & \Rightarrow x^2-2x+1=0\\ & \Rightarrow x=1\Rightarrow y=2\end{array}$$

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 QR:

$$x_2+x_3=2x_1=2,y_2+y_3=2y_1=4</p><p>$$So, we can write: $x_3=2-x_2,y_3=4-y_2$

Given,

$$PQ=\frac{2\sqrt{2}}{3}\Rightarrow P{Q}^2={\left(x_2-1\right)}^2+{\left(y_2-2\right)}^2=\frac{8}{9}$$

Let's denote: $x_2=a,y_2=b,x_3=2-a,y_3=4-b$

$$\begin{array}{rl}& (a-1{)}^2+(b-2{)}^2=\frac{8}{9}\\ & \Rightarrow a^2-2a+1+b^2-4b+4=\frac{8}{9}\\ & \Rightarrow a^2+b^2-2a-4b+5=\frac{8}{9}\\ & \Rightarrow 9a^2+9b^2-18a-36b+37=0\end{array}$$

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

$$\begin{array}{rl}& x_1{y}_1+x_2{y}_2+x_3{y}_3=2+ab+(2-a)(4-b)\\ & 9\left(x_1{y}_1+x_2{y}_2+x_3{y}_3\right)=9(10+2ab-2b-4a)\\ & =90+18ab-18b-36a=46\end{array}$$