• # question_answer The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, -3), then its radius is : [JEE Main 10-4-2019 Morning] A) $3\sqrt{2}$B) $3$C) $2\sqrt{2}$                D) $2$

Equation of circle can be written as ${{\left( x1 \right)}^{2}}+{{\left( y1 \right)}^{2}}+\lambda \left( xy \right)=0$ It passes through (1, -3) $16+\lambda \left( 4 \right)=0\Rightarrow \lambda =-4$ So ${{\left( x1 \right)}^{2}}+{{\left( y1 \right)}^{2}}4\left( xy \right)=0$ $\Rightarrow {{x}^{2}}+{{y}^{2}}6x+2y+2=0$ $\Rightarrow r=2\sqrt{2}$ (correct key is b)