JEE Main & Advanced
JEE Main Paper (Held on 10-4-2019 Morning)
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)\[2\]
Correct Answer:
A
Solution :
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)