Question

Write the standard form of the equation and the general form of the equation of each circle of radius \(r\) and center \((h, k)\). Graph each circle. $$ r=\sqrt{13} ;(h, k)=(5,-1) $$

Short answer

Standard form: \((x-5)^2 + (y+1)^2 = 13\); General form: \((x^2 - 10x + y^2 + 2y + 13 = 0)\).

Step-by-step solution

Write the Standard Form of the Circle's Equation

The standard form of the equation of a circle with radius \(r\) and center \((h, k)\) is \((x-h)^2 + (y-k)^2 = r^2\). Given \(r=\sqrt{13}\) and center \((h, k) = (5, -1)\), substitute these values into the formula.\[(x-5)^2 + (y+1)^2 = (\sqrt{13})^2\]Simplify the equation:\[(x-5)^2 + (y+1)^2 = 13\]

Write the General Form of the Circle's Equation

To write the general form of the equation \((x-5)^2 + (y+1)^2 = 13\) , expand and simplify.First, expand \((x-5)^2\) and \((y+1)^2\):\[ (x-5)^2 = x^2 - 10x + 25\]\[ (y+1)^2 = y^2 + 2y + 1\]Now, combine and simplify:\[ x^2 - 10x + 25 + y^2 + 2y + 1 = 13\]Combine the constants:\[ x^2 - 10x + y^2 + 2y + 26 - 13 = 0\]\[ x^2 - 10x + y^2 + 2y + 13 = 0 \]Thus, the general form of the circle's equation is \((x^2 - 10x + y^2 + 2y + 13 = 0)\).

Graph the Circle

To graph the circle, plot the center at \((5, -1)\) and draw a circle with the given radius \((r = \sqrt{13} \approx 3.61)\). You'll need to make sure the distance from the center to any point on the circle is approximately 3.61 units.

Key concepts

Standard Form of Circle Equation

The standard form of a circle's equation helps us define any circle using its radius and center. The standard form is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center, and \(r\) is the radius. Let's break it down to understand better. If you know the center and radius, you can easily write the equation.
For example, given a circle with radius \(r=\sqrt{13}\) and center at \((5, -1)\), we can substitute these values into the formula:
\[(x-5)^2 + (y+1)^2 = (\root{13})^2\]
Simplify, and you get:
\[(x-5)^2 + (y+1)^2 = 13\].

• \(x-5\) represents the horizontal distance from the center
• \(y+1\) represents the vertical distance from the center
• Together, these squared terms equal the radius squared

You only need to substitute values for your specific problem, and this standard form provides a straightforward way to describe any circle.

General Form of Circle Equation

Once you have the standard form of an equation, you can convert it into the general form. The general form of a circle's equation is written as \(Ax^2 + Ay^2 + Dx + Ey + F = 0\). It gives a complete polynomial expression to define the circle. Let’s see how to transform a standard form into the general form.
Using the standard form equation we have, \[(x-5)^2 + (y+1)^2 = 13\], we first expand \((x-5)^2\) and \((y+1)^2\):
\( (x-5)^2 = x^2 - 10x + 25 \)
\( (y+1)^2 = y^2 + 2y + 1 \)
Combine and simplify:
\( x^2 - 10x + 25 + y^2 + 2y + 1 = 13 \)
Finally, bring all terms to one side to generalize:
\( x^2 - 10x + y^2 + 2y + 26 - 13 = 0 \)
which simplifies to:
\( x^2 - 10x + y^2 + 2y + 13 = 0 \)
Now you have the circle's equation in the general form.

Graphing Circles

Graphing a circle involves plotting its center and radius on a coordinate plane. First, find the center from the standard form equation and plot it on the graph. For the given circle, our center is at \((5, -1)\).
Next, use the radius to determine the distance from the center to points on the circle. For the circle we've worked with, the radius is \(\sqrt{13} \approx 3.61\) units.
Here’s how to graph it:
1. Plot the center at \((5, -1)\).
2. From the center, measure \(3.61\) units in all directions (up, down, left, right, and diagonally).
3. Draw a smooth curve connecting these points equidistant from the center.

• Use a compass or drawing tool to ensure accuracy.
• Ensure that the circle remains symmetric around the center.

Now you have a visual representation of your circle, making it easier to understand and analyze any related problems.