jee-main 2024 Q76

jee-main · India · session2_06apr_shift2 Differential equations Solving Separable DEs with Initial Conditions

Suppose the solution of the differential equation $\frac { d y } { d x } = \frac { ( 2 + \alpha ) x - \beta y + 2 } { \beta x - 2 \alpha y - ( \beta \gamma - 4 \alpha ) }$ represents a circle passing through origin. Then the radius of this circle is:
(1) 2
(2) $\sqrt { 17 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { \sqrt { 17 } } { 2 }$

Suppose the solution of the differential equation $\frac { d y } { d x } = \frac { ( 2 + \alpha ) x - \beta y + 2 } { \beta x - 2 \alpha y - ( \beta \gamma - 4 \alpha ) }$ represents a circle passing through origin. Then the radius of this circle is:\\
(1) 2\\
(2) $\sqrt { 17 }$\\
(3) $\frac { 1 } { 2 }$\\
(4) $\frac { \sqrt { 17 } } { 2 }$

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