The order of the differential equation of all circles whose radius is 4 , is

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Accepted Answer

Correct answer: B. (b) : Equation of a circle having radius =4 is (x-a )^2+ (y-b )^2=4^2 where (a ,b ) is the centre of the circle. ⇒ (x-a )^2+ (y-b )^2=16 On differentiating w.r.t. ' x ', we get 2 (x-a )+2 (y-b )y^ '=0 ⇒ (x-a )+ (y-b )y^ '=0 On differentiating again w.r.t. ' x ', we get 1+ (y^ ' )^2+ (y-b )y^ ' '=0 ⇒y-b= -(1+ (y^ ' )^2)y^ ' ' Substituting value of (y-b ) in (ii), we get (x-a )- (1+ (y^ ' )^2)y^ ' 'y^ '=0 ⇒ (x-a )= (1+ (y^ ' )^2)y^ 'y^ ' ' Now, substituting value of (x-a ) and (y-b ) in (i), we get ( (1+ (y^ ' )^2)y^ 'y^ ' ' )^2+ ( -(1+ (y^ ' )^2)y^ ' ' )^2=4^2 ⇒ ((1+ (y^ ' )^2)y^ ' )^2+(1+ (y^ ' )^2)^2=16 (y^ ' ' )^2 ∴ Order is 2

$The order of the differential equation of all circles whose radius is 4 , is$

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