Question

Find the differential equation of all the circles which pass through the origin and whose centers lie on the x - 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 x - axis, so k = 0

So, we have the equation of circle as, (x – h)² + y² = r²

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

0 + h² = r²

So, the equation of circle is (x – h)² + y² = h²

⇒ x² – 2hx + y² = 0

⇒ x² + y² = 2hx

Now, differentiating it with respect to x we get,

Hence, the required differential equation is \((x^2-y^2)+2xy\frac{dy}{dx}=0\)