Question

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

Answer 1

To find: differential equation of all circles which pass through origin and centre lies on y axis

Assume a point (0, k) on the y axis

Therefore, the radius of circle is given as

And the general form of equation of circle is

(x-a)²+(y-b)²=r²

Where (a, b) is the centre and r is radius, now substituting the value in the above equation we get

(x-0)²+(y-k)²=k²

x²+y²-2yk=0 …. (i)

Differentiating both side with respect to x