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 presents a differential equation whose solution is a circle that passes through the origin. The objective is to determine the radius of this circle by first deriving its equation from the provided differential equation.

Concept Used:

The solution approach involves identifying and solving an exact differential equation and then utilizing the geometric properties of a circle.

1. A differential equation in the form \( M(x, y)dx + N(x, y)dy = 0 \) is classified as exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
2. The general solution for an exact differential equation is formulated as: \[ \int M \, dx \text{ (treating y as a constant)} + \int (\text{terms in N that do not contain x}) \, dy = C \]
3. The general equation representing a second-degree curve is \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \). For this equation to define a circle, two specific conditions must be satisfied:
   • (a) The coefficient of the \( xy \) term must be zero, meaning \( h = 0 \).
   • (b) The coefficients of \( x^2 \) and \( y^2 \) must be equal, meaning \( a = b \).
4. For a circle represented by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), the radius \( R \) is calculated using the formula \( R = \sqrt{g^2 + f^2 - c} \).

Step-by-Step Solution:

Step 1: The given differential equation is rewritten into the standard form \( M(x, y)dx + N(x, y)dy = 0 \).

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

Upon cross-multiplication, the equation becomes:

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

Rearranging the terms results in:

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

This equation conforms to the \( M dx + N dy = 0 \) format, 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: The exactness of the equation is verified.

An equation is considered exact if the partial derivative of \( M \) with respect to \( y \) equals the partial derivative of \( N \) with respect to \( x \), i.e., \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

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

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

Step 3: The general solution for the exact differential equation is derived.

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

Multiplying the entire equation by 2 to eliminate the fractions yields:

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

Letting \( C' = 2C \), the equation is expressed as:

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

Step 4: The conditions necessary for the equation to represent a circle are applied.

Condition (a) dictates that the coefficient of \( xy \) must be zero:

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

Condition (b) requires that the coefficients of \( x^2 \) and \( y^2 \) be equal:

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

Step 5: The determined values of \( \alpha \) and \( \beta \) are substituted into the solution equation.

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

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

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

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

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

Consequently, the equation of the circle is:

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

Dividing by 4 standardizes the equation:

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

Step 7: The radius of the circle is computed.

Comparing \( x^2 + y^2 + x - 4y = 0 \) with the general circle equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we identify the coefficients:

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

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

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

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