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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