Suppose the solution of the differential equation $\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}$ represents a circle passing through origin. Then the radius of this circle is : Options: (A) 2 (B) $\sqrt{17}$ (C) $\frac{1}{2}$ (D) $\frac{\sqrt{17}}{2}$

### Analyzing the Setup The given differential equation is: $$\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}$$ To begin, we perform cross-multiplication to transform this into the standard differential form $M dx + N dy = 0$. This yields the following expression: $$((\beta x - 2\alpha y - (\beta\gamma - 4\alpha)) dy = ((2+\alpha)x - \beta y + 2) dx$$ Rearranging all terms to one side, we obtain our primary battlefield: $$((2+\alpha)x - \beta y + 2) dx - (\beta x - 2\alpha y - \beta\gamma + 4\alpha) dy = 0$$ ### The 'Aha!' Moment We now look for the underlying structure of the equation by grouping terms strategically. We identify the term $-\beta(y dx + x dy)$, which is the **exact differential** of the product $xy$. Recall that $d(xy) = x dy + y dx$. Recognizing this identity allows us to simplify the expression significantly. ### The Path to Integration With the terms grouped, we integrate each part individually: $$\int (2+\alpha)x dx + \int 2 dx + \int 2\alpha y dy + \int (\beta\gamma - 4\alpha) dy - \beta \int d(xy) = C$$ Performing the integration, we arrive at the general solution: $$\frac{(2+\alpha)x^2}{2} + 2x + \alpha y^2 + (\beta\gamma - 4\alpha)y - \beta xy = C$$ ### The Geometry of the Circle For this second-degree equation to represent a **circle**, two specific conditions must be satisfied: 1. The coefficient of the $xy$ term must be $0$. 2. The coefficients of $x^2$ and $y^2$ must be equal. Applying the first condition, the coefficient of $xy$ is $-\beta$, which implies $\beta = 0$. Applying the second condition, we equate the coefficients of $x^2$ and $y^2$: $$\frac{2+\alpha}{2} = \alpha \Rightarrow 2+\alpha = 2\alpha \Rightarrow \alpha = 2$$ ### The Final Reveal Given $\alpha = 2$ and $\beta = 0$, and knowing the circle passes through the origin $(0,0)$, we substitute these values to find $C = 0$. The equation of the curve becomes: $$2x^2 + 2y^2 + 2x - 8y = 0$$ Dividing by $2$, we obtain the standard form: $$x^2 + y^2 + x - 4y = 0$$ Comparing this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify the center $(-\frac{1}{2}, 2)$. The radius $r$ is calculated as: $$r = \sqrt{g^2 + f^2 - c} = \sqrt{\left(\frac{1}{2}\right)^2 + (-2)^2 - 0} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{17}{4}}$$ The final radius of the circle is **$\frac{\sqrt{17}}{2}$**.

JEE Main 2024 (06 April Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Suppose the solution of the differential equation $\frac{d y}{d x} = \frac{( 2 + α ) x - β y + 2}{β x - 2 α y - ( β γ - 4 α )}$ represents a circle passing through origin. Then the radius of this circle is :

Q: Why does the circle passing through the origin make $C=0$?

The constant $C$ is the result of the integration process. When we say the circle passes through the origin $(0,0)$, it means that the point $(0,0)$ must satisfy the equation of the circle. By substituting $x=0$ and $y=0$ into our integrated equation, all terms involving $x$ and $y$ vanish, leaving us with $0 = C$. It is a very elegant way to determine the constant!

Select Answer:

Progressive Hints

Hint 1

The differential equation looks intimidating, but it is a classic case of a non-homogeneous equation that can be rearranged into the form M dx + N dy = 0. Your first task is to cross-multiply and group the terms. Look for a structure that resembles the product rule of differentiation, specifically

d (x y) = x d y + y d x

Hint 2

Once you have integrated the equation, you will have a general second-degree equation in

x

and

y

. Recall the necessary conditions for a general second-degree equation

A x^{2} + B x y + C y^{2} + D x + E y + F = 0

to represent a circle: the coefficient of the

x y

term must be zero (

B = 0

), and the coefficients of

x^{2}

and

y^{2}

must be equal (

A = C

Hint 3

After determining the parameters

α

and

β

using the circle conditions, use the fact that the circle passes through the origin

(0, 0)

to find the constant of integration

C

. Finally, write the equation in the standard form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

and use the radius formula

r = g^{2} + f^{2} - c

Visualized Solution

The Given Differential Equation

Given Differential Equation: $\frac{d y}{d x} = \frac{( 2 + α ) x - β y + 2}{β x - 2 α y - ( β γ - 4 α )}$
Goal: Find the radius of the circle represented by its solution.
Constraint: The circle passes through the origin $(0, 0)$ .

Rearranging to Differential Form

Cross-multiply to form $M d x + N d y = 0$ :
$(β x - 2 α y - (β γ - 4 α)) d y = ((2 + α) x - β y + 2) d x$
Rearranging terms:
$((2 + α) x - β y + 2) d x - (β x - 2 α y - β γ + 4 α) d y = 0$

Grouping for Exact Differentials

Expand and group terms strategically:
$(2 + α) x d x + 2 d x + 2 α y d y + (β γ - 4 α) d y - β (y d x + x d y) = 0$
Recognize the exact differential:
$d (x y) = x d y + y d x$

Integrating the Equation

Integrate the equation term by term:
$\int (2 + α) x d x + \int 2 d x + \int 2 α y d y + \int (β γ - 4 α) d y - β \int d (x y) = C$
Resulting general solution:
$(2 + α) \frac{x ^{2}}{2} + 2 x + α y^{2} + (β γ - 4 α) y - β x y = C$

Conditions for a Circle

For a general second-degree equation to represent a circle:
1. The coefficient of $x y$ must be exactly $0$ .
2. The coefficient of $x^{2}$ must equal the coefficient of $y^{2}$ .

Applying Circle Condition 1

Apply Condition 1 (No $x y$ term):
Coefficient of $x y = - β$
$- β = 0 \Rightarrow β = 0$

Applying Circle Condition 2

Apply Condition 2 (Equal squared coefficients):
Coefficient of $x^{2} = \frac{2 + α}{2}$
Coefficient of $y^{2} = α$
$\frac{2 + α}{2} = α \Rightarrow 2 + α = 2 α \Rightarrow α = 2$

Passing Through the Origin

Apply the origin constraint $(0, 0)$ :
Substitute $x = 0, y = 0$ into the integrated equation:
$(2 + 2) \frac{0 ^{2}}{2} + 2 (0) + 2 (0)^{2} + (0 \cdot γ - 4 \cdot 2) (0) = C$
$0 = C \Rightarrow C = 0$

The Final Circle Equation

Substitute $α = 2, β = 0, C = 0$ back into the equation:
$\frac{2 + 2}{2} x^{2} + 2 x + 2 y^{2} + (0 - 4 (2)) y = 0$
$2 x^{2} + 2 y^{2} + 2 x - 8 y = 0$
Divide by $2$ to get the standard circle equation:
$x^{2} + y^{2} + x - 4 y = 0$

Finding Center and Radius

Compare with standard form: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$2 g = 1 \Rightarrow g = \frac{1}{2}$
$2 f = - 4 \Rightarrow f = - 2$
$c = 0$
Center $(- g, - f) = (- \frac{1}{2}, 2)$

Calculating the Radius

Radius formula: $r = g^{2} + f^{2} - c$
$r = (\frac{1}{2})^{2} + (- 2)^{2} - 0$
$r = \frac{1}{4} + 4 = \frac{17}{4}$
$r = \frac{17}{2}$

Final Conclusion

Final Conclusion:
The radius of the circle is $\frac{17}{2}$ .
This matches option (4).

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} = \frac{( 2 + α ) x - β y + 2}{β x - 2 α y - ( β γ - 4 α )}

To begin, we perform cross-multiplication to transform this into the standard differential form

M d x + N d y = 0

This yields the following expression:

((β x - 2 α y - (β γ - 4 α)) d y = ((2 + α) x - β y + 2) d x

Rearranging all terms to one side, we obtain our primary battlefield:

((2 + α) x - β y + 2) d x - (β x - 2 α y - β γ + 4 α) d y = 0

The 'Aha!' Moment

We now look for the underlying structure of the equation by grouping terms strategically. We identify the term

- β (y d x + x d y)

, which is the exact differential of the product

x y

Recall that

d (x y) = x d y + y d x

. Recognizing this identity allows us to simplify the expression significantly.

The Path to Integration

With the terms grouped, we integrate each part individually:

\int (2 + α) x d x + \int 2 d x + \int 2 α y d y + \int (β γ - 4 α) d y - β \int d (x y) = C

Performing the integration, we arrive at the general solution:

\frac{( 2 + α ) x ^{2}}{2} + 2 x + α y^{2} + (β γ - 4 α) y - β x y = C

The Geometry of the Circle

For this second-degree equation to represent a circle, two specific conditions must be satisfied:

1. The coefficient of the

x y

term must be

0

. 2. The coefficients of

x^{2}

and

y^{2}

must be equal.

Applying the first condition, the coefficient of

x y

- β

, which implies

β = 0

Applying the second condition, we equate the coefficients of

x^{2}

and

y^{2}

\frac{2 + α}{2} = α \Rightarrow 2 + α = 2 α \Rightarrow α = 2

The Final Reveal

Given

α = 2

and

β = 0

, and knowing the circle passes through the origin

(0, 0)

, we substitute these values to find

C = 0

The equation of the curve becomes:

2 x^{2} + 2 y^{2} + 2 x - 8 y = 0

Dividing by

2

, we obtain the standard form:

x^{2} + y^{2} + x - 4 y = 0

Comparing this to the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the center

(- \frac{1}{2}, 2)

The radius

r

is calculated as:

r = g^{2} + f^{2} - c = (\frac{1}{2})^{2} + (- 2)^{2} - 0 = \frac{1}{4} + 4 = \frac{17}{4}

The final radius of the circle is $\frac{17}{2}$ .

Common Traps & FAQs

Wait, why can't we just solve this as a homogeneous differential equation?

That is a great observation! A differential equation is homogeneous if the numerator and denominator are both homogeneous functions of the same degree. Here, the presence of the constant terms

2

and

- (β γ - 4 α)

breaks that symmetry. Because of these constants, the equation is not homogeneous, and we must look for other patterns, like exact differentials.

How did you know

d (x y) = x d y + y d x

was hidden in the equation?

It comes down to pattern recognition. When you see terms like

y d x

and

x d y

appearing together, it is almost always a hint to look for the product rule of differentiation. By grouping the terms carefully, we can transform a complex expression into a simple, integrable form. Always keep an eye out for these 'exact differentials' in JEE problems!

Why does the circle passing through the origin make

C = 0

The constant

C

is the result of the integration process. When we say the circle passes through the origin

(0, 0)

, it means that the point

(0, 0)

must satisfy the equation of the circle. By substituting

x = 0

and

y = 0

into our integrated equation, all terms involving

x

and

y

vanish, leaving us with

0 = C

. It is a very elegant way to determine the constant!

Animated Solution for Mathematics - Differential Equations: Suppose the solution of the differential equation $\frac{d y}{d x} = \frac{( 2 + α ) x - β y + 2}{β x - 2 α y - ( β γ - 4 α )}$ represents a circle passing through origin. Then the radius of this circle is :

Select Answer:

Visualized Solution

The Given Differential Equation

Rearranging to Differential Form

Grouping for Exact Differentials

Integrating the Equation

Conditions for a Circle

Applying Circle Condition 1

Applying Circle Condition 2

Passing Through the Origin

The Final Circle Equation

Finding Center and Radius

Calculating the Radius

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The 'Aha!' Moment

The Path to Integration

The Geometry of the Circle

The Final Reveal

Similar Questions

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Select Answer:

Visualized Solution

The Given Differential Equation

Rearranging to Differential Form

Grouping for Exact Differentials

Integrating the Equation

Conditions for a Circle

Applying Circle Condition 1

Applying Circle Condition 2

Passing Through the Origin

The Final Circle Equation

Finding Center and Radius

Calculating the Radius

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The 'Aha!' Moment

The Path to Integration

The Geometry of the Circle

The Final Reveal

Similar Questions

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