jee-main 2023 Q68
Circle Equation Derivation
The points of intersection of the line $a x + b y = 0 , ( \mathrm { a } \neq \mathrm { b } )$ and the circle $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } - 2 \mathrm { x } = 0$ are $A ( \alpha , 0 )$ and $B ( 1 , \beta )$. The image of the circle with $A B$ as a diameter in the line $\mathrm { x } + \mathrm { y } + 2 = 0$ is:
(1) $x ^ { 2 } + y ^ { 2 } + 5 x + 5 y + 12 = 0$
(2) $x ^ { 2 } + y ^ { 2 } + 3 x + 5 y + 8 = 0$
(3) $x ^ { 2 } + y ^ { 2 } + 3 x + 3 y + 4 = 0$
(4) $x ^ { 2 } + y ^ { 2 } - 5 x - 5 y + 12 = 0$