# Form the differential equation of all the circles which pass through the origin and whose centers lie on the y – axis.

Any circle with centre at (h, k) and radius r is given by,

(x – h)2 + (y – k)2 = r2

Here centre is on y – axis, so h = 0

So, we have the equation of circle as, x2 + (y – k)2 = r2

Further, it is given that circle passes through the origin (0,0) therefore origin must satisfy the equation of circle. So, we get,

0 + k2 = r2

So, the equation of circle is x2 + (y – k)2 = k2

x2 + y2 – 2ky = 0

x2 + y2 = 2ky Now, differentiating it with respect to x we get,     Hence, the required differential equation is Rate this question :

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