Problem 67 - Algebra and Trigonometry

Question

\(65-72\) . Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}-4 x+10 y+13=0 $$

Step-by-Step Solution

Answer

The circle's center is (2, -5) and the radius is 4.

Step 1: Identify the General Form

The given equation is \(x^{2} + y^{2} - 4x + 10y + 13 = 0\). Identify its terms: quadratic terms are \(x^{2}\) and \(y^{2}\); linear terms are \(-4x\) and \(10y\); and the constant is \(13\).

Step 2: Group and Complete the Square for x

Focus on the terms involving \(x\): \(x^{2} - 4x\). To complete the square, take the coefficient of \(x\), which is \(-4\), divide by 2 to get \(-2\), and square it to add \(4\). This transforms \(x^{2} - 4x\) into \((x - 2)^{2}\). Subtract \(4\) to maintain equality.

Step 3: Group and Complete the Square for y

Now focus on the terms involving \(y\): \(y^{2} + 10y\). Take the coefficient of \(y\), which is \(10\), divide by 2 to get \(5\), and square it to add \(25\). This transforms \(y^{2} + 10y\) into \((y + 5)^{2}\). Subtract \(25\) to keep the equation balanced.

Step 4: Rewrite the Equation in Circle Form

Update the equation with the completed squares: \((x - 2)^{2} - 4 + (y + 5)^{2} - 25 + 13 = 0\). Simplify the constant terms: \(-4 - 25 + 13 = -16\). Hence, the equation becomes \((x - 2)^{2} + (y + 5)^{2} = 16\).

Step 5: Identify Center and Radius

The circle's equation is \((x - h)^{2} + (y - k)^{2} = r^{2}\). From \((x - 2)^{2} + (y + 5)^{2} = 16\), the center \((h, k)\) is \((2, -5)\) and the radius \(r\) is \(\sqrt{16} = 4\).

Key Concepts

Completing the Square

Completing the square is a key mathematical method that helps in rewriting quadratic expressions to reveal their characteristics more clearly. Imagine you have any terms such as \(x^2 - 4x\). Here, our goal is to transform these terms into a perfect square trinomial. Why do we do this? Because perfect squares are easier to work with and give us valuable insights.

To complete the square for a term like \(x^2 - 4x\), the method is simple and systematic:

• Take the coefficient of \(x\), which is \(-4\).
• Divide it by 2, yielding \(-2\).
• Square this result to get \(4\).

Now we've constructed a perfect square trinomial \((x - 2)^2\), correct? But remember, we adjusted the original equation when we added \(4\). To maintain the balance, we subtract \(4\) as well. This method is also applied to terms involving \(y\), such as \(y^2 + 10y\), where we ultimately form \((y + 5)^2\). Completing the square helps uncover the direct form of the equation of a circle.

Center of a Circle

The center of a circle in coordinate geometry is vital as it serves as the anchor point from which every point on the circle maintains a constant distance (the radius). When dealing with circle equations in the format \((x - h)^2 + (y - k)^2 = r^2\), the coordinates \((h, k)\) represent the circle's center.

In our specific example, after completing the square, we derived the equation \((x - 2)^2 + (y + 5)^2 = 16\). From this, we can see the values \(h = 2\) and \(k = -5\). Hence, the center of this circle is at \((2, -5)\).
It is important to remember that the expressions \(h\) and \(-k\) in the equation tell us exactly how much we "shift" the basic circle \((x)^2 + (y)^2 = r^2\) from the origin, making graphing and understanding its position straightforward.

Radius of a Circle

Finding the radius of a circle using its equation is straightforward yet crucial.
The equation of a circle often appears in the form \((x - h)^2 + (y - k)^2 = r^2\), where \(r^2\) is the radius squared. Knowing \(r^2\) allows us to easily calculate the radius, \(r\), by taking the square root.

In our processed form of the circle equation \((x - 2)^2 + (y + 5)^2 = 16\), the term \(16\) represents \(r^2\). Therefore, the radius \(r\) is the square root: \(\sqrt{16} = 4\).

This tells us that every point on the circle sits exactly 4 units away from its center. Understanding the radius is crucial not just for graphing, but also for solving real-life problems involving circles where knowing a fixed distance from the center is important.