Q. 14
Form the differential equation of all circles which pass through origin and whose centres lie on y-axis
Answer :
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
Formula:
Substituting the value of k in eq(i)
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