Question

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

Answer 1

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 (x² - y²)\(\frac{dy}{dx}=2xy\)