Finding the Equation of a Circle

The question asks for the equation of a circle given its center and radius. This is a standard problem in coordinate geometry, specifically dealing with conic sections.

Standard Equation of a Circle

The standard form of the equation of a circle with center $(\text{h}, \text{k})$ and radius $\text{r}$ is given by:

\((x - h)^2 + (y - k)^2 = r^2\)

In this form, the coordinates of the center are easily identifiable, and the radius is the square root of the constant on the right side.

Applying the Given Information

We are given the following information for the circle:

• Center \((h, k) = (1, -2)\)
• Radius \(r = 4\) cm

Now, we substitute these values into the standard equation of a circle:

\((x - 1)^2 + (y - (-2))^2 = 4^2\)

Simplify the term \(y - (-2)\) to \(y + 2\), and \(4^2\) to \(16\):

\((x - 1)^2 + (y + 2)^2 = 16\)

Expanding the Equation

To match the format of the given options, we need to expand the squared terms and rearrange the equation.

Expand \((x - 1)^2\) using the formula \((a - b)^2 = a^2 - 2ab + b^2\):

\((x - 1)^2 = x^2 - 2(x)(1) + 1^2 = x^2 - 2x + 1\)

Expand \((y + 2)^2\) using the formula \((a + b)^2 = a^2 + 2ab + b^2\):

\((y + 2)^2 = y^2 + 2(y)(2) + 2^2 = y^2 + 4y + 4\)

Substitute these expanded forms back into the equation:

\((x^2 - 2x + 1) + (y^2 + 4y + 4) = 16\)

Rearranging to General Form

Combine like terms and move the constant to the right side to get the general form of the equation of the circle.

\(x^2 + y^2 - 2x + 4y + 1 + 4 = 16\)

\(x^2 + y^2 - 2x + 4y + 5 = 16\)

Subtract 5 from both sides of the equation:

\(x^2 + y^2 - 2x + 4y = 16 - 5\)

\(x^2 + y^2 - 2x + 4y = 11\)

This is the equation of the circle with the given center and radius.

Verification by Comparison

Let's compare our derived equation with the given options:

• Option 1: \(x^2 + y^2 + 2x - 4y = 16\) (Incorrect coefficients for x and y, incorrect constant)
• Option 2: \(x^2 + y^2 + 2x - 4y = 11\) (Incorrect coefficients for x and y)
• Option 3: \(x^2 + y^2 + 2x + 4y = 16\) (Incorrect coefficient for x, incorrect constant)
• Option 4: \(x^2 + y^2 - 2x + 4y = 11\) (Matches our derived equation)

Therefore, the correct equation of the circle is \(x^2 + y^2 - 2x + 4y = 11\).

Step-by-Step Solution Summary

1. Identify the standard equation of a circle: \((x - h)^2 + (y - k)^2 = r^2\).
2. Substitute the given center \((h, k) = (1, -2)\) and radius \(r = 4\) into the standard equation.
3. Simplify the equation: \((x - 1)^2 + (y + 2)^2 = 16\).
4. Expand the squared terms: \((x^2 - 2x + 1) + (y^2 + 4y + 4) = 16\).
5. Combine constants and rearrange to get the general form: \(x^2 + y^2 - 2x + 4y + 5 = 16\).
6. Move the constant to the right side: \(x^2 + y^2 - 2x + 4y = 11\).
7. Compare the final equation with the given options to find the match.

Revision Table: Equation of Circle

Additional Information: Circle Geometry

A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center. The distance from the center to any point on the circle is called the radius.

Understanding the standard form \((x - h)^2 + (y - k)^2 = r^2\) is crucial as it directly involves the center and radius, making it easy to write the equation when these parameters are known, or to find these parameters when the equation is given in this form.

The general form \(x^2 + y^2 + Dx + Ey + F = 0\) is obtained by expanding the standard form. From the general form, one can find the center and radius by using the formulas or by completing the square for the x and y terms.

For the equation \(x^2 + y^2 - 2x + 4y - 11 = 0\), \(D = -2\), \(E = 4\), and \(F = -11\). Using the formulas:

• Center: \((-D/2, -E/2) = (-(-2)/2, -(4)/2) = (2/2, -4/2) = (1, -2)\)
• Radius squared: \((D/2)^2 + (E/2)^2 - F = (-2/2)^2 + (4/2)^2 - (-11) = (-1)^2 + (2)^2 + 11 = 1 + 4 + 11 = 16\)
• Radius: \(\sqrt{16} = 4\)

This confirms that our derived equation correctly represents a circle with center (1, -2) and radius 4.