Question

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 the origin. Then the radius of this circle is:

• A. $\sqrt{17}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{17}}{2}$
• D. 2

Answer 1

The Correct Option is C

The problem provides a differential equation whose solution is a circle passing through the origin. We need to find the radius of this circle by first determining its equation from the given differential equation.

Concept Used:

The solution involves recognizing and solving an exact differential equation and applying the geometric properties of a circle.

1. A differential equation of the form \( M(x, y)dx + N(x, y)dy = 0 \) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
2. The general solution of an exact differential equation is given by: \[ \int M \, dx \text{ (treating y as constant)} + \int (\text{terms in N not containing x}) \, dy = C \]
3. The general equation of a second-degree curve is \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). For this to represent a circle, two conditions must be met:
   • (a) The coefficient of the \( xy \) term must be zero, i.e., \( h = 0 \).
   • (b) The coefficients of \( x^2 \) and \( y^2 \) must be equal, i.e., \( a = b \).
4. The radius \( R \) of a circle with the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by the formula \( R = \sqrt{g^2 + f^2 - c} \).

Step-by-Step Solution:

Step 1: Rewrite the given differential equation in the standard form \( M(x, y)dx + N(x, y)dy = 0 \).

\[ \frac{dy}{dx} = \frac{(2 + \alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta \gamma - 4\alpha)} \]

Cross-multiplying gives:

\[ (\beta x - 2\alpha y - (\beta \gamma - 4\alpha)) dy = ((2 + \alpha)x - \beta y + 2) dx \]

Rearranging the terms, we get:

\[ ((2 + \alpha)x - \beta y + 2) dx - (\beta x - 2\alpha y - (\beta \gamma - 4\alpha)) dy = 0 \]

This is in the form \( M dx + N dy = 0 \), where:

\[ M = (2 + \alpha)x - \beta y + 2 \] \[ N = -(\beta x - 2\alpha y - (\beta \gamma - 4\alpha)) = -\beta x + 2\alpha y + (\beta \gamma - 4\alpha) \]

Step 2: Check if the equation is exact.

An equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}((2 + \alpha)x - \beta y + 2) = -\beta \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-\beta x + 2\alpha y + (\beta \gamma - 4\alpha)) = -\beta \]

Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the differential equation is exact. Its solution will yield the equation of the curve.

Step 3: Find the general solution of the exact differential equation.

\[ \int M \, dx + \int (\text{terms in N not containing x}) \, dy = C \] \[ \int ((2 + \alpha)x - \beta y + 2) \, dx + \int (2\alpha y + (\beta \gamma - 4\alpha)) \, dy = C \] \[ \frac{(2 + \alpha)x^2}{2} - \beta xy + 2x + \frac{2\alpha y^2}{2} + (\beta \gamma - 4\alpha)y = C \]

Multiplying the entire equation by 2 to clear the fraction, we get:

\[ (2 + \alpha)x^2 - 2\beta xy + 4x + 2\alpha y^2 + 2(\beta \gamma - 4\alpha)y = 2C \]

Let \( C' = 2C \). The equation is:

\[ (2 + \alpha)x^2 + 2\alpha y^2 - 2\beta xy + 4x + 2(\beta \gamma - 4\alpha)y - C' = 0 \]

Step 4: Apply the conditions for the equation to represent a circle.

Condition (a): The coefficient of \( xy \) must be zero.

\[ -2\beta = 0 \implies \beta = 0 \]

Condition (b): The coefficients of \( x^2 \) and \( y^2 \) must be equal.

\[ 2 + \alpha = 2\alpha \implies \alpha = 2 \]

Step 5: Substitute the values of \( \alpha \) and \( \beta \) into the solution equation.

With \( \alpha = 2 \) and \( \beta = 0 \), the equation becomes:

\[ (2 + 2)x^2 + 2(2)y^2 - 2(0)xy + 4x + 2(0 \cdot \gamma - 4 \cdot 2)y - C' = 0 \] \[ 4x^2 + 4y^2 + 4x - 16y - C' = 0 \]

Step 6: Use the condition that the circle passes through the origin \( (0, 0) \).

Substituting \( x = 0 \) and \( y = 0 \) into the equation:

\[ 4(0)^2 + 4(0)^2 + 4(0) - 16(0) - C' = 0 \implies C' = 0 \]

So, the equation of the circle is:

\[ 4x^2 + 4y^2 + 4x - 16y = 0 \]

Dividing by 4, we get the standard form:

\[ x^2 + y^2 + x - 4y = 0 \]

Step 7: Calculate the radius of the circle.

Comparing \( x^2 + y^2 + x - 4y = 0 \) with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we have:

\[ 2g = 1 \implies g = \frac{1}{2} \] \[ 2f = -4 \implies f = -2 \] \[ c = 0 \]

The formula for the radius is \( R = \sqrt{g^2 + f^2 - c} \).

\[ R = \sqrt{\left(\frac{1}{2}\right)^2 + (-2)^2 - 0} \] \[ R = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{1 + 16}{4}} = \sqrt{\frac{17}{4}} \] \[ R = \frac{\sqrt{17}}{2} \]

The radius of the circle is \( \frac{\sqrt{17}}{2} \).

Answer 2

\[ \frac{dy}{dx} = \frac{(2 + \alpha)x - \beta y + 2}{\beta x - y(2\alpha + \beta) + 4\alpha} \]

\[ \beta xdy - (2\alpha + \beta)ydy + 4\alpha dy = (2 + \alpha)x dx - \beta ydx + 2dx \]

\[ \beta(xdy + ydx) - (2\alpha + \beta)ydy + 4\alpha dy = (2 + \alpha)x dx + 2dx \]

\[ \beta xy - \frac{(2\alpha + \beta)y^2}{2} + 4\alpha y = \frac{(2 + \alpha)x^2}{2} \]

\[ \implies \beta = 0 \quad \text{for this to be a circle} \]

\[ (2 + \alpha)\frac{x^2}{2} + \alpha y^2 + 2x - 4xy = 0 \]

\[ \text{Coeff. of } x^2 = y^2 \implies 2 + a = 2a \implies \alpha = 2 \]

\[ \text{i.e., } 2x^2 + 2y^2 + 2x - 8y = 0 \]

\[ x^2 + y^2 + x - 4y = 0 \]

\[ \text{rd} = \sqrt{\frac{1}{4} + 4} = \frac{\sqrt{17}}{2} \]