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 equation represents a circle centered at (2, -5) with a radius of 4.

Step 1: Identify the Circle Equation

A circle's standard equation is \((x-h)^2 + (y-k)^2 = r^2\). The given equation is \(x^2 + y^2 - 4x + 10y + 13 = 0\). First, we need to rewrite it in the circle's standard form by completing the square for both the \(x\) and \(y\) terms.

Step 2: Rearrange the Equation

Rearrange the terms: \(x^2 - 4x + y^2 + 10y = -13\). This will set up the equation for us to complete the square.

Step 3: Complete the Square for x

Take the terms involving \(x\): \(x^2 - 4x\). Complete the square by adding and subtracting \(4\) (which is \((\frac{-4}{2})^2\)) to get the expression \((x - 2)^2 - 4\).

Step 4: Complete the Square for y

Take the terms involving \(y\): \(y^2 + 10y\). Complete the square by adding and subtracting \(25\) (which is \((\frac{10}{2})^2\)) to get the expression \((y + 5)^2 - 25\).

Step 5: Substitute Back and Simplify

Re-substitute the completed squares back into the equation: \((x - 2)^2 - 4 + (y + 5)^2 - 25 = -13\). Simplify it to \((x - 2)^2 + (y + 5)^2 = 16\).

Step 6: Identify the Center and Radius

From the equation \((x-2)^2 + (y+5)^2 = 16\), we see it represents a circle centered at \((2, -5)\) with radius \(r\) where \(r^2 = 16\). Therefore, \(r = 4\).

Key Concepts

Completing the Square

Completing the square is a technique that transforms quadratic equations into a form that is easier to work with. It is particularly useful when trying to rewrite an equation into a standard form. This method involves creating a perfect square trinomial from a quadratic expression, such as \(x^2 - 4x\).

Here's how you do it:

• Identify the coefficient of the linear term, which is \(-4\) in \(x^2 - 4x\).
• Divide this coefficient by 2, which gives you \(-2\).
• Square the result \((-2)^2 = 4\) and add and subtract it within the expression to maintain equality.

This results in \((x - 2)^2 - 4\). Notice that \((x - 2)^2\) is a perfect square.

Repeating these steps for \(y^2 + 10y\) would involve squaring \(\frac{10}{2} = 5\) to add and subtract 25, which gives \((y + 5)^2 - 25\). Completing the square streamlines complex equations, setting the stage for finding the standard form of a circle.

Standard Form of a Circle

The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\). It offers a concise way to express a circle in terms of its center \((h, k)\) and radius \(r\). To convert an equation like \(x^2 + y^2 - 4x + 10y + 13 = 0\) into this form, rearrange terms and complete the square.

Once you complete the square for both the \(x\) and \(y\) components, you substitute back into the original equation. The equation now simplifies to \((x - 2)^2 + (y + 5)^2 = 16\) after moving constants and completing the square. This equation matches the standard form where \((x - h)^2\) and \((y - k)^2\) relate directly to the circle's center coordinates, and \(r^2\) provides the radius squared. The standard form makes it straightforward to identify a circle's key characteristics.

Equation of a Circle

The equation of a circle in geometry is central to understanding circles within the coordinate plane. The general equation \(x^2 + y^2 - 4x + 10y + 13 = 0\) initially seems complex. However, by rewriting it in the standard form, we can easily understand its properties.

Through completing the square, we transform this equation into \((x - 2)^2 + (y + 5)^2 = 16\), clearly indicating a geometric circle. The equation reflects the balance and symmetry inherent in a circle: equal distance (radius) from the center \((h, k)\) to any point on the circle. The left side of the simplified equation represents both horizontal and vertical components squared to reflect this distance.

Center and Radius of a Circle

Identifying the center and radius of a circle from its equation is straightforward once it is expressed in the standard form \((x-h)^2 + (y-k)^2 = r^2\).

For the equation \((x - 2)^2 + (y + 5)^2 = 16\):

• The circle's center is at \((h, k)\), found from the expressions \((x-2)^2\) and \((y+5)^2\), indicating the center \((2, -5)\).
• The radius \(r\) is derived by taking the square root of the constant on the right side. In this case, since \(r^2 = 16\), the radius \(r = 4\).

Identifying these properties helps in visualizing the circle and understanding its position and size within the coordinate plane. The center is the balance point, and the radius determines the extent of the circle.