JEE Main 2025 — Circle Question with Solution

$$\begin{array}{rl}& x-y+1=0\\ & \mathrm{p}=\mathrm{r}\\ & \left|\frac{\alpha -0+1}{\sqrt{2}}\right|=r\Rightarrow (\alpha +1{)}^2=2r^2\text{.... (1)}\end{array}$$

\begin{aligned} & \text { now }\left(\frac{-3 \alpha+0-1}{\sqrt{9+4}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2=\mathrm{r}^2 \\ & \Rightarrow(3 \alpha+1)^2+4=13 \mathrm{r}^2 \ldots \ldots .(2) \\ & \text { (1) & }(2) \Rightarrow(3 \alpha+1)^2+4=13 \frac{(\alpha+1)^2}{2} \\ & \quad \Rightarrow 18 \alpha^2+12 \alpha+2+8=13 \alpha^2+26 \alpha+13 \\ & \Rightarrow 5 \alpha^2-14 \alpha-3=0 \\ & \Rightarrow 5 \alpha^2-15 \alpha+\alpha-3=0 \\ & \Rightarrow 5 \alpha^2-15 \alpha+\alpha-3=0 \\ & \Rightarrow \alpha=\frac{-1}{5}, 3 \end{aligned}

$$\begin{array}{rl}& \therefore r=2\sqrt{2}\\ & \text{ How }\alpha \mathrm{e}=3\text{ and }2\alpha =4\sqrt{2}\\ & \begin{array}{rl}& {\alpha }^2{\mathrm{e}}^2=9\Rightarrow \alpha =2\sqrt{2}\Rightarrow {\alpha }^2=8\\ & {\alpha }^2\left(1+\frac{{\beta }^2}{{\alpha }^2}\right)=9\\ & {\alpha }^2+{\beta }^2=9\\ & \therefore {\beta }^2=1\\ & \therefore 2{\alpha }^2+3{\beta }^2=2(8)+3(1)=19\end{array}\end{array}$$