Determining the center and radius of a circle from its equation is a fundamental concept in geometry. While the standard form of a circle's equation, $(x-h)^2 + (y-k)^2 = r^2$, directly reveals the center $(h,k)$ and radius $r$, many times we encounter circles represented by a general degree 2 equation. This article will delve into the process of finding the center and radius of a circle given a general degree 2 equation, illustrating the methodology with clear examples.
Understanding the General Degree 2 Equation
A general degree 2 equation representing a circle takes the form:
$Ax^2 + Ay^2 + Dx + Ey + F = 0$
where A, D, E, and F are constants. The key observation is that the coefficients of $x^2$ and $y^2$ are equal, a condition that ensures the equation describes a circle.
Converting to Standard Form
To extract the center and radius, we need to transform the general equation into the standard form. This process involves completing the square for both the $x$ and $y$ terms:
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Group the $x$ and $y$ terms:
Rearrange the equation to group the $x$ terms together and the $y$ terms together:
$Ax^2 + Dx + Ay^2 + Ey = -F$
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Factor out the coefficient of $x^2$ and $y^2$:
$A(x^2 + \frac{D}{A}x) + A(y^2 + \frac{E}{A}y) = -F$
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Complete the square for the $x$ terms:
- Take half of the coefficient of the $x$ term ($\frac{D}{2A}$), square it ($\frac{D^2}{4A^2}$), and add it inside the parentheses on the left-hand side.
- Since we are adding $\frac{D^2}{4A}$ to the left-hand side, we must also add it to the right-hand side to maintain the equation's balance.
$A(x^2 + \frac{D}{A}x + \frac{D^2}{4A^2}) + A(y^2 + \frac{E}{A}y) = -F + \frac{D^2}{4A}$
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Complete the square for the $y$ terms:
- Follow the same procedure as for the $x$ terms: take half of the coefficient of the $y$ term ($\frac{E}{2A}$), square it ($\frac{E^2}{4A^2}$), and add it inside the parentheses on the left-hand side.
- Add $\frac{E^2}{4A}$ to the right-hand side to maintain equality.
$A(x^2 + \frac{D}{A}x + \frac{D^2}{4A^2}) + A(y^2 + \frac{E}{A}y + \frac{E^2}{4A^2}) = -F + \frac{D^2}{4A} + \frac{E^2}{4A}$
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Rewrite the left-hand side as squares:
$A(x + \frac{D}{2A})^2 + A(y + \frac{E}{2A})^2 = -F + \frac{D^2}{4A} + \frac{E^2}{4A}$
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Divide both sides by $A$ to isolate the squared terms:
$(x + \frac{D}{2A})^2 + (y + \frac{E}{2A})^2 = \frac{-F + \frac{D^2}{4A} + \frac{E^2}{4A}}{A}$
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Simplify the right-hand side:
$(x + \frac{D}{2A})^2 + (y + \frac{E}{2A})^2 = \frac{-4AF + D^2 + E^2}{4A^2}$
Now the equation is in the standard form:
$(x - h)^2 + (y - k)^2 = r^2$
where:
- Center: $(h, k) = (-\frac{D}{2A}, -\frac{E}{2A})$
- Radius: $r = \sqrt{\frac{-4AF + D^2 + E^2}{4A^2}}$
Example: Finding the Center and Radius
Let's illustrate the process with an example. Consider the equation:
$2x^2 + 2y^2 - 8x + 12y - 24 = 0$
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Group the terms:
$2x^2 - 8x + 2y^2 + 12y = 24$
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Factor out the coefficient of $x^2$ and $y^2$:
$2(x^2 - 4x) + 2(y^2 + 6y) = 24$
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Complete the square for $x$:
$2(x^2 - 4x + 4) + 2(y^2 + 6y) = 24 + 8$
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Complete the square for $y$:
$2(x^2 - 4x + 4) + 2(y^2 + 6y + 9) = 24 + 8 + 18$
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Rewrite as squares:
$2(x - 2)^2 + 2(y + 3)^2 = 50$
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Divide by 2:
$(x - 2)^2 + (y + 3)^2 = 25$
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Identify the center and radius:
- Center: $(h, k) = (2, -3)$
- Radius: $r = \sqrt{25} = 5$
Therefore, the circle represented by the equation $2x^2 + 2y^2 - 8x + 12y - 24 = 0$ has a center at $(2, -3)$ and a radius of $5$.
Conclusion
Finding the center and radius of a circle given a general degree 2 equation involves converting the equation to standard form through the process of completing the square. By following these steps, we can easily determine the center and radius, which are essential parameters for understanding and analyzing circles. This process is crucial for various applications in geometry, calculus, and other areas of mathematics and physics.
