MHT CET · Maths · Differential Equations

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

A $(x^{2} + y^{2}) \frac{d y}{d x} - 2 x y = 0$
B $x^{2} - y^{2} + 2 x y \frac{d y}{d x} = 0$
C $(x^{2} - y^{2}) \frac{d y}{d x} - 2 x y = 0$
D $(x^{2} + y^{2}) \frac{d y}{d x} + 2 x y = 0$

Verified Solution

Answer & Solution

Correct Answer

(B) $x^{2} - y^{2} + 2 x y \frac{d y}{d x} = 0$

Step-by-step Solution

Detailed explanation

As required circle touches y - axis at the origin.

∴

Let Center of the circle is

d (a, 0)

and radius is a

∴

Equation of circle will be,

{(x - a)}^{2} + {(y - 0)}^{2} = a^{2}

\Rightarrow x^{2} - 2 a x + a^{2} + y^{2} = a^{2}

\Rightarrow x^{2} + y^{2} - 2 a x = 0

.... (i)
By differentiating above equation w.r.t x, we get

2 x + 2 y \frac{d y}{d x} - 2 a = 0

∴ 2 a = 2 x + 2 y \frac{d y}{d x}

.... (ii)
From (i) and (ii),

x^{2} {+ y}^{2} - (2 x + 2 y \frac{d y}{d x}) x = 0

\Rightarrow x^{2} + y^{2} - 2 x^{2} - 2 x y \frac{d y}{d x} = 0

\Rightarrow x^{2} - y^{2} + 2 x y \frac{d y}{d x} = 0

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app