The differential equation of family of circles with fixed radius 5 units and centre on the line y = 2, is

$(y-2)^2y'^2=25-(y-2)^2$

The correct answer is option (1) : $(y-2)^2y'^2=25-(y-2)^2$

Let (a, 2) be the centre of the circle, where 'a' is a variable. Then, the equation of teh family of cirlces is

$(x-a)^2 +(y-2)^2 =5^2$ .............(i)

Differentiating w.r. to x, we get

$2(x-a) + 2(y-2) \frac{dy}{dx}⇒x-a=-(y-2)y_1$

Substitute this value of (x-a) in (i), we get

$(y-2)^2 y_1^2= 25- (y-2)^2$

as the required differential equation.