Problem 11 - Contemporary Precalculus

Question

Find the center and radius of the circle whose equation is given. $$x^{2}+y^{2}+25 x+10 y=-12$$

Step-by-Step Solution

Answer

Answer: The center of the circle is \(\left(-\frac{25}{2}, -5\right)\), and the radius is \(\sqrt{143.25}\).

Step 1: Group the x and y terms

In this step, we group the \(x\) and \(y\) terms together. Rearrange the given equation: $$x^{2}+25x+y^{2}+10y=-12$$

Step 2: Complete the square for x terms

Next, we complete the square for the x terms. Take half of the coefficient of the \(x\) term (which is \(25\)), square it (\(\frac{25}{2} \times \frac{25}{2} = 156.25\)) and then add and subtract it from the equation: $$\left(x^{2}+25x+\frac{25^2}{2^2}-\frac{25^2}{2^2}\right)+y^{2}+10y=-12$$

Step 3: Complete the square for y terms

Now, complete the square for the y terms. Take half of the coefficient of the \(y\) term (which is \(10\)), square it (\(\frac{10}{2} \times \frac{10}{2} = 25\)) and then add and subtract it from the equation: $$\left(x^{2}+25x+\frac{25^2}{2^2}-\frac{25^2}{2^2}\right)+\left(y^{2}+10y+\frac{10^2}{2^2}-\frac{10^2}{2^2}\right)=-12$$

Step 4: Rewrite the equation in standard form

Now rewrite the equation in the standard form of a circle equation. Notice the appropriate terms that will collapse into squares: $$\left(x+\frac{25}{2}\right)^2 -\left(\frac{25}{2}\right)^2 + \left(y+\frac{10}{2}\right)^2 -\left(\frac{10}{2}\right)^2 = -12$$ Simplify: $$\left(x+\frac{25}{2}\right)^2 - 156.25 + \left(y+5\right)^2 - 25 = -12$$ Now, bring the constants to the right side: $$\left(x+\frac{25}{2}\right)^2 + \left(y+5\right)^2 = 156.25 - 25 + 12$$ Which simplifies to: $$\left(x+\frac{25}{2}\right)^2 + \left(y+5\right)^2 = 143.25$$

Step 5: Identify the center and radius of the circle

From the standard form we obtained, we can identify the center \((a, b)\) and the radius \(r\) of the circle. We can see that the center is at: $$\left(-\frac{25}{2}, -5\right)$$ And the radius is: $$r = \sqrt{143.25}$$

Key Concepts

Completing the Square

To find the center and radius of a circle from its equation, we first convert the given equation into a suitable format. This involves the method known as "completing the square."

Completing the square helps to simplify a quadratic expression into a binomial square. This is advantageous because it converts the circle equation into a format that's easy to identify the center and radius. In our example, we tackle the expression that contains the \(x\) terms and separately the \(y\) terms.

• For the \(x\) terms \(x^2 + 25x\), take half of the linear coefficient (25), square it \((\frac{25}{2})^2 = 156.25\), and add and subtract it.
• For the \(y\) terms \(y^2 + 10y\), similarly, take half of the linear coefficient (10), square it \((\frac{10}{2})^2 = 25\), then add and subtract it.

By adding the squared result to the equation, you ensure that you are creating a perfect square trinomial, which rewrites the equation into an easier-to-read form. This is a crucial step to simplify the flow of the circle equation into its standard form.

Circle Center

Once the equation of a circle is in its standard form, it is much easier to identify the circle's properties. The standard form of a circle's equation is \((x - a)^2 + (y - b)^2 = r^2\), where \((a, b)\) is the center.

From our worked example, after completing the square, the equation becomes \((x + \frac{25}{2})^2 + (y + 5)^2 = 143.25\). The format reveals the modified center coordinates in the equation:

• Since it is \((x + \frac{25}{2})^2\), the center's x-coordinate is -\(\frac{25}{2}\).
• Since it is \((y + 5)^2\), the center's y-coordinate is -5.

Notice that from these values, the variables in the binomials have signs opposite to that of the actual center coordinates because of the equation's structure \((x - a)\) form, hence the negative values for both coordinates in the found center.

Circle Radius

The radius of a circle can be easily determined once you have the equation in the standard form \((x - a)^2 + (y - b)^2 = r^2\). Here, \(r\) represents the radius.

In our equation example, the right-hand side already represents \(r^2\). To find \(r\), simply take the square root of this value. Therefore, from the equation \(143.25\), we deduce the following:

• The radius \(r\) is the square root of 143.25.

Utilizing a calculator or suitable software can find the numeric value for the radius, resulting in \(r \approx \sqrt{143.25}\).

Understanding this step is crucial for not only interpreting the geometric size of the circle but for further applications where precise measurements are needed. The radius, being a measure of distance from the center to the circle, quantifies the circle's size.