Suppose the solution of the differential equation d x d y = β x − 2 α y − ( β γ − 4 α ) ( 2 + α ) x − β y + 2 represents a circle passing through origin. Then the radius of this circle is :

Question

Suppose the solution of the differential equation d x d y ​ = β x − 2 α y − ( β γ − 4 α ) ( 2 + α ) x − β y + 2 ​ represents a circle passing through origin. Then the radius of this circle is :

TestHub · Accepted Answer

​ d x d y ​ = β x − y ( 2 α + β ) + 4 α ( 2 + α ) x − β y + 2 ​ β x d y − ( 2 α + β ) y d y + 4 α d y = ( 2 + α ) x d x − β y d x + 2 d x β ( x d y + y d x ) − ( 2 α + β ) y d y + 4 α d y = ( 2 + α ) x d x + 2 d x β x y − 2 ( 2 α + β ) y 2 ​ + 4 α y = 2 ( 2 + α ) x 2 ​ ​ ​ ⇒ β = 0 for this to be circle ( 2 + α ) 2 x 2 ​ + α y 2 + 2 x − 4 α y = 0 ​ ​ i.e. 2 x 2 + 2 y 2 + 2 x − 8 y = 0 x 2 + y 2 + x − 4 y = 0 r d = 4 1 ​ + 4 ​ = 2 17 ​ ​ ​

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