Circle Equation Transform To Standard Form, Identify Center And Radius

In the realm of analytic geometry, circles hold a fundamental position. Their equations, particularly in the standard form, provide valuable insights into their geometric properties, such as their center and radius. This article delves into the process of transforming a general equation of a circle into the standard form $(x-h)^2+(y-k)^2=c$, where $(h, k)$ represents the center of the circle, and $c$ is related to the radius. We will also explore scenarios where the equation might represent a degenerate case, and how to identify and express their solution sets. Let's consider the given equation $x^2+y^2-16x+6y+9=0$ and embark on the journey of converting it to the standard form.

Completing the Square: The Key to Transformation

The core technique to transform the given equation $x^2+y^2-16x+6y+9=0$ into the standard form involves a method called completing the square. This method allows us to rewrite quadratic expressions as perfect squares, which are essential for obtaining the desired form $(x-h)^2+(y-k)^2=c$. To begin, we group the $x$ terms and $y$ terms together and move the constant term to the right side of the equation:

$(x^2-16x) + (y^2+6y) = -9$

Now, we focus on completing the square for the $x$ terms and the $y$ terms separately. For the $x$ terms, we take half of the coefficient of the $x$ term (-16), which is -8, square it (-8)^2 = 64, and add it to both sides of the equation. Similarly, for the $y$ terms, we take half of the coefficient of the $y$ term (6), which is 3, square it (3)^2 = 9, and add it to both sides of the equation. This process ensures that we maintain the equality of the equation while creating perfect square trinomials:

$(x^2-16x+64) + (y^2+6y+9) = -9 + 64 + 9$

Now, we can rewrite the expressions in parentheses as perfect squares:

$(x-8)^2 + (y+3)^2 = 64$

Identifying the Center and Radius

The equation is now in the standard form $(x-h)^2+(y-k)^2=c$. By comparing our transformed equation $(x-8)^2 + (y+3)^2 = 64$ with the standard form, we can readily identify the center and radius of the circle. The center of the circle is given by the coordinates $(h, k)$, where $h$ is the value subtracted from $x$ inside the first parenthesis, and $k$ is the value subtracted from $y$ inside the second parenthesis. In our case, $h = 8$ and $k = -3$ (since we have $y + 3$, which can be written as $y - (-3)$). Therefore, the center of the circle is $(8, -3)$.

The value $c$ on the right side of the equation represents the square of the radius. In our case, $c = 64$. To find the radius, we take the square root of $c$: radius = $\sqrt{64} = 8$. Hence, the radius of the circle is 8 units.

In summary, by completing the square, we transformed the original equation into the standard form $(x-8)^2 + (y+3)^2 = 64$, which revealed that the equation represents a circle with a center at $(8, -3)$ and a radius of 8.

Exploring Degenerate Cases

While many equations of the form $x^2 + y^2 + Dx + Ey + F = 0$ represent circles, there are cases where they might represent a degenerate case. A degenerate case occurs when the equation does not describe a typical circle. This can happen in a few scenarios, which we will explore further.

One degenerate case arises when the value of $c$ in the standard form $(x-h)^2+(y-k)^2=c$ is equal to zero. In this situation, the equation becomes $(x-h)^2+(y-k)^2=0$. The only solution to this equation is when both $(x-h)^2$ and $(y-k)^2$ are equal to zero simultaneously. This implies that $x = h$ and $y = k$. Geometrically, this represents a single point at the coordinates $(h, k)$, rather than a circle with a non-zero radius. The solution set in this case consists of just one point, which can be written as ${(h, k)}$.

Another possible degenerate case occurs when the value of $c$ is negative. If $c$ is negative, the equation $(x-h)^2+(y-k)^2=c$ has no real solutions. This is because the sum of two squares, $(x-h)^2$ and $(y-k)^2$, can never be negative for real values of $x$, $y$, $h$, and $k$. Therefore, the solution set in this case is the empty set, denoted by $\emptyset$, indicating that there are no points that satisfy the equation.

To determine if an equation represents a degenerate case, it is crucial to transform it into the standard form $(x-h)^2+(y-k)^2=c$ using the method of completing the square. Once in standard form, the value of $c$ will reveal whether the equation represents a typical circle (when $c > 0$), a single point (when $c = 0$), or no solution (when $c < 0$).

Examples of Degenerate Cases

To solidify our understanding of degenerate cases, let's examine a couple of examples.

Example 1: The Single Point Case

Consider the equation $x^2 + y^2 - 4x + 6y + 13 = 0$. To determine if it represents a degenerate case, we complete the square:

$(x^2 - 4x) + (y^2 + 6y) = -13$

$(x^2 - 4x + 4) + (y^2 + 6y + 9) = -13 + 4 + 9$

$(x - 2)^2 + (y + 3)^2 = 0$

In this case, $c = 0$. Therefore, the equation represents a single point at $(2, -3)$. The solution set is ${(2, -3)}$.

Example 2: The Empty Set Case

Now, let's analyze the equation $x^2 + y^2 + 2x - 8y + 20 = 0$. Completing the square gives us:

$(x^2 + 2x) + (y^2 - 8y) = -20$

$(x^2 + 2x + 1) + (y^2 - 8y + 16) = -20 + 1 + 16$

$(x + 1)^2 + (y - 4)^2 = -3$

Here, $c = -3$, which is negative. This indicates that the equation has no real solutions, and the solution set is the empty set, $\emptyset$.

These examples illustrate how the value of $c$ in the standard form dictates the nature of the solution set, differentiating between a typical circle, a single point, and an empty set.

Conclusion

Transforming the general equation of a circle into the standard form $(x-h)^2+(y-k)^2=c$ is a powerful technique that unveils the circle's center and radius. The method of completing the square is instrumental in this transformation. Furthermore, understanding degenerate cases, where the equation represents a single point or no solution, provides a comprehensive understanding of the various forms a circle equation can take. By analyzing the value of $c$ in the standard form, we can effectively determine the nature of the solution set and gain valuable insights into the geometric representation of the equation. The ability to manipulate and interpret circle equations is a fundamental skill in analytic geometry, with applications spanning diverse fields such as physics, engineering, and computer graphics.