Question

Find equation of smallest circle passing from points(1,1) and (2,2) which is always in the first quadrant

Answer 1

$$\begin{gathered} \mathrm{That}\mathrm{circle}\mathrm{will}\mathrm{be}\mathrm{taking}\left(1,1\right),\left(2,2\right)\mathrm{as}\mathrm{diameter}\mathrm{so} \\ \left(\mathrm{x}-{\mathrm{x}}_1\right)\left(\mathrm{x}-{\mathrm{x}}_2\right)+\left(\mathrm{y}-{\mathrm{y}}_1\right)\left(\mathrm{y}-{\mathrm{y}}_2\right)=0 \\ \left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)+\left(\mathrm{y}-1\right)\left(\mathrm{y}-2\right)=0 \\ {\mathrm{x}}^2-3\mathrm{x}+2+{\mathrm{y}}^2-3\mathrm{y}+2=0 \\ {\mathrm{x}}^2+{\mathrm{y}}^2-3\mathrm{x}-3\mathrm{y}+4=0 \end{gathered}$$