Single Choice

The differential equation of the family of circles touching y-axis at the origin is:

A$$\left( { x }^{ 2 }+{ y }^{ 2 } \right) \cfrac { dy }{ dx } -2xy=0$$
B$${ x }^{ 2 }-{ y }^{ 2 }+2xy\cfrac { dy }{ dx } =0$$
Correct Answer
C$$\left( { x }^{ 2 }-{ y }^{ 2 } \right) \cfrac { dy }{ dx } -2xy=0$$
D$$\left( { x }^{ 2 }+{ y }^{ 2 } \right) \cfrac { dy }{ dx } +2xy=0$$

Solution

The system of circles touching Y axis at origin will 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 $$a$$
$$(x ─ a)² + (y ─ 0)² = a²$$
That 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 equation of system of circles touching Y axis at origin, eliminate the the arbitrary constant from equation(1)
Differentiating equation(1) with respect to x,
$$2x + 2y dy/dx ─ 2a=0$$
or
$$2a = 2(x + y dy/dx)$$
Replacing '2a' of equation(1) with the above expression, you get
$$x² + y² ─ 2(x + y dy/dx)(x) = 0$$
That is,
$$─x² + y² ─2xy dy/dx = 0$$
or
$$x² ─ y² + 2xy dy/dx = 0$$

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