Login

form the differential equation of a family of circles touching y- axis at origin

The system of circles touching Y axis at origin will have centres on X axis.
Let (a,0) be the centre of a circle as the y coordinate is zero.
Then the radius of the circle should be a units,
Equation of a circle with centre at (a,0) and radius a:
(x ─ a)² + (y ─ 0)² = a²
Or x2 + a2 -2ax + y2 = a2
So x2  -2ax + y2 = 0
Here there is one arbitrary constant, so we have to differentiate it once to eliminate a.
2x -2a +2ydydx = 0So 2a =2x + 2ydydxHence putting in the initial equation, we have x2 -(2x + 2ydydx)x + y2 = 0-x2-2xydydx +y2 =0 So it is the differential form required.

  • 25
View Full Answer

plz ans this fast

  • 3

The system of circlestouching Y axis at originwill have centres on X axis. Let (a,0) be the centre of a circle. Then the radius of the circle should be a units, since the circle should touch Y axis at origin.Equation of a circle with centre at (a,0) and radius a is(x ─ a)² + (y ─ 0)² = a²That is,x² + y² ─ 2ax = 0 ─────► (1)The above equation represents the family of circles touching Y axis at origin. Here 'a' is an arbitrary constant.In order to find the differential equationof system of circles touching Y axis at origin, eliminate the the arbitrary constant from equation(1)Differentiating equation(1) with respect to x,2x + 2y dy/dx ─ 2a = 0or2a = 2(x + y dy/dx)Replacing '2a' of equation(1) with the above expression, you getx² + y² ─ 2(x + y dy/dx)(x) = 0That is,─x² + y² ─2xy dy/dx = 0orx² ─ y² + 2xy dy/dx = 0

  • 5
What are you looking for?