The general solution of $\frac{{dy}}{{dx}} = \frac{{ax + h}}{{by + k}}$ represents a circle only when

Question

Prepp Team · Accepted Answer

Understanding the Differential Equation and Conic Sections

The given differential equation is:

$\frac{{dy}}{{dx}} = \frac{{ax + h}}{{by + k}}$

We are asked to find the condition under which the general solution of this differential equation represents a circle. To do this, we first need to solve the differential equation to find its general solution.

Solving the Differential Equation

We can solve this first-order differential equation by separating variables:

$(by + k) \, dy = (ax + h) \, dx$

Now, we integrate both sides of the equation:

$\int (by + k) \, dy = \int (ax + h) \, dx$

Integrating gives:

$\frac{b y^2}{2} + ky = \frac{a x^2}{2} + hx + C$

where $C$ is the constant of integration.

Rearranging into the General Form of a Conic Section

Let's rearrange the terms to get the equation into the general form of a conic section, which is typically written as $Ax^2 + By^2 + Cx + Dy + E = 0$:

$\frac{a x^2}{2} - \frac{b y^2}{2} + hx - ky + C = 0$

Multiplying the entire equation by 2 to remove fractions (and letting $2C$ be a new constant $C'$ for simplicity):

$a x^2 - b y^2 + 2hx - 2ky + C' = 0$

This is the general equation of a conic section, where the coefficients are $A = a$, $B = -b$, $C_{gen} = 2h$, $D_{gen} = -2k$, and $E = C'$.

Condition for a Circle

For a general conic section equation of the form $Ax^2 + By^2 + Cx_{gen} + Dy_{gen} + E = 0$ to represent a circle, two main conditions must be satisfied:

The coefficients of the $x^2$ term and the $y^2$ term must be equal ($A = B$).
These equal coefficients must be non-zero ($A = B \ne 0$).
There must be no $xy$ term (which is already satisfied in our derived equation).

Applying the Circle Condition

Comparing our derived equation $a x^2 - b y^2 + 2hx - 2ky + C' = 0$ with the general conic section form, we have $A = a$ and $B = -b$.

Applying the conditions for a circle:

$A = B \implies a = -b$

And

$A \ne 0 \implies a \ne 0$

Since $a = -b$, if $a \ne 0$, then $-b \ne 0$, which means $b \ne 0$.

Combining these, the condition for the solution to represent a circle is $a = -b$ and $a \ne 0$. This can be written compactly as $a = -b \ne 0$.

Analyzing the Options

Let's examine the given options based on the derived condition:

Option 1: a = b = 0 If $a = 0$ and $b = 0$, the original differential equation becomes $\frac{{dy}}{{dx}} = \frac{{h}}{{k}}$. The solution is $ky = hx + C$, which is the equation of a straight line, not a circle.
Option 2: a = -b $\ne$ 0 This matches the condition we derived for the general solution to be a circle. If $a = -b \ne 0$, the equation becomes $a x^2 - (-a) y^2 + 2hx - 2ky + C' = 0$, which simplifies to $a x^2 + a y^2 + 2hx - 2ky + C' = 0$. Since $a \ne 0$, we can divide by $a$: $x^2 + y^2 + \frac{2h}{a}x - \frac{2k}{a}y + \frac{C'}{a} = 0$. This is the standard form of a circle equation: $(x - g)^2 + (y - f)^2 = r^2$, where center is $(-g, -f)$ and radius is $r$, provided $g^2 + f^2 - \frac{C'}{a} > 0$. The coefficients of $x^2$ and $y^2$ are equal (both 1 after dividing by $a$) and non-zero.
Option 3: a = b $\ne$ 0, h = k If $a = b \ne 0$, the equation is $a x^2 - a y^2 + 2hx - 2ky + C' = 0$. This is $a(x^2 - y^2) + 2hx - 2ky + C' = 0$. Since the coefficients of $x^2$ ($a$) and $y^2$ ($-a$) are equal in magnitude but opposite in sign ($a \ne -a$ unless $a=0$, but we are given $a \ne 0$), this represents a hyperbola. The condition $h=k$ affects the position of the center but not the type of conic section.
Option 4: a = b $\ne$ 0 As explained in option 3, if $a = b \ne 0$, the equation represents a hyperbola.

Based on the analysis, the condition $a = -b \ne 0$ ensures that the general solution of the given differential equation represents a circle.

Condition	Equation Form	Type of Conic Section
$a = -b \ne 0$	$a x^2 + a y^2 + 2hx - 2ky + C' = 0$	Circle (provided discriminant is positive)
$a = b \ne 0$	$a x^2 - a y^2 + 2hx - 2ky + C' = 0$	Hyperbola
$a = 0, b = 0$	$2hx - 2ky + C' = 0$	Straight Line

Revision Table: Differential Equation and Circle Condition

Concept	Description	Relevance to Question
Differential Equation	An equation involving derivatives of a function.	The starting point, requires solving.
Separation of Variables	A method to solve certain first-order differential equations by separating variables on opposite sides.	Method used to solve the given DE.
General Solution	The solution to a differential equation that includes an arbitrary constant.	The equation whose form we analyze.
Conic Section	A curve formed by the intersection of a plane and a double cone (circle, ellipse, parabola, hyperbola). Represented by $Ax^2 + By^2 + Cx + Dy + E = 0$.	The class of curves the general solution belongs to.
Condition for Circle	For $Ax^2 + By^2 + Cx + Dy + E = 0$, a circle requires $A=B \ne 0$.	The specific condition applied to determine when the solution is a circle.

Additional Information: Conic Section Identification

The general second-degree equation in two variables is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. The type of conic section it represents depends on the discriminant $\Delta = B^2 - 4AC$.

If $\Delta < 0$: Ellipse (or a circle if $A=C$ and $B=0$, or a point, or no locus if degenerate).
If $\Delta = 0$: Parabola (or a pair of parallel lines, coinciding lines, or no locus if degenerate).
If $\Delta > 0$: Hyperbola (or a pair of intersecting lines if degenerate).

In our derived equation $a x^2 - b y^2 + 2hx - 2ky + C' = 0$, we have $A=a$, $B=0$, and $C=-b$. The discriminant is $\Delta = 0^2 - 4(a)(-b) = 4ab$.

For a circle, we need $a = -b \ne 0$. In this case, $4ab = 4a(-a) = -4a^2$. Since $a \ne 0$, $a^2 > 0$, so $-4a^2 < 0$. The discriminant is negative, which correctly identifies it as an ellipse (a circle is a special case of an ellipse where $A=C$).
For a hyperbola (when $a = b \ne 0$), $4ab = 4a(a) = 4a^2$. Since $a \ne 0$, $4a^2 > 0$. The discriminant is positive, correctly identifying it as a hyperbola.

This confirms that our condition $a = -b \ne 0$ leads to the correct classification based on the discriminant as well.

Prepp Team · Answer

a = b = 0

Prepp Team · Answer

a = b &ne; 0, h = k

Prepp Team · Answer

a = b &ne; 0