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. $$ r=4 ;(h, k)=(-2,1) $$

Short answer

\((x + 2)^2 + (y - 1)^2 = 16\); \(x^2 + y^2 + 4x - 2y - 11 = 0\)

Step-by-step solution

Identify given parameters

Given the radius of the circle is \(r = 4\) and the center of the circle \((h, k) = (-2, 1)\).

Write the standard form of the circle's equation

The standard form of the circle's equation is given by \((x - h)^2 + (y - k)^2 = r^2\). Substituting \(h = -2\), \(k = 1\), and \(r = 4\): \((x - (-2))^2 + (y - 1)^2 = 4^2\) which simplifies to: \((x + 2)^2 + (y - 1)^2 = 16\).

Write the general form of the circle's equation

To write the general form, expand the standard form equation: \((x + 2)^2 + (y - 1)^2 = 16 \)Expanding each term, \( (x + 2)^2 = x^2 + 4x + 4 \) and \( (y - 1)^2 = y^2 - 2y + 1 \). Substitute back to get: \( x^2 + 4x + 4 + y^2 - 2y + 1 = 16 \), Combine and rearrange all terms to one side: \( x^2 + y^2 + 4x - 2y + 5 - 16 = 0 \). Thus, the general form is: \( x^2 + y^2 + 4x - 2y - 11 = 0 \).

Graph the circle using the given parameters

To graph the circle, plot the center at \((-2, 1)\) and draw a circle with radius 4 units around it.

Key concepts

standard form of a circle

The standard form of a circle's equation is a fundamental concept in geometry. It originates from the distance formula, which measures the distance between two points in a plane. The standard form is expressed as \( (x - h)^2 + (y - k)^2 = r^2 \). Here, \( (h, k) \) represents the center of the circle and \( r \) is the radius.
To illustrate this with an example, let's use the given circle parameters: center \( (h, k) = (-2, 1) \) and radius \( r = 4 \). Plugging these values into the standard form, we get:
\[ (x + 2)^2 + (y - 1)^2 = 16 \]
This equation tells us that any point \( (x, y) \) lying on the circle is 4 units away from its center at \( (-2, 1) \).
Understanding this form is crucial as it simplifies the process of identifying a circle's key features such as its center and radius from an equation.
Next, we'll transform this standard form into the general form of a circle for further insights.

general form of a circle

The general form of a circle's equation is derived by expanding and simplifying the standard form. It is presented as \( Ax^2 + By^2 + Cx + Dy + E = 0 \). For a circle, the coefficients of \( x^2 \) and \( y^2 \) are equal and positive, making \( A = B \).
Using the same example, we start with the standard form:
\[ (x + 2)^2 + (y - 1)^2 = 16 \]
Expand the binomials:
\[ x^2 + 4x + 4 + y^2 - 2y + 1 = 16 \]
Combine like terms and move the constant to the other side:
\[ x^2 + y^2 + 4x - 2y - 11 = 0 \]
The final form, \( x^2 + y^2 + 4x - 2y - 11 = 0 \), is the general form of the given circle's equation.
This form is useful for identifying whether a given quadratic equation represents a circle and determining its center and radius by completing the square.
Now that we have both forms, let’s see how to graph this circle.

graphing circles

Graphing circles involves locating the center and plotting points at the radius distance. From the given circle parameters (center \( -2, 1 \) and radius \( 4 \)), follow these steps:
* Plot the center at \((-2, 1)\).
* From the center, measure 4 units outward in all directions to identify points on the perimeter. Points like \((-6, 1)\), \( (2, 1) \), \( (-2, -3) \), and \( (-2, 5) \) are crucial as they represent the radius distance in the horizontal and vertical directions.
* Draw a smooth curve connecting these perimeter points to form the circle.
To check your work, any point on this circle should satisfy its equation \( (x + 2)^2 + (y - 1)^2 = 16 \). For instance, substituting \( x = -6 \) and \( y = 1 \) yields \((-6 + 2)^2 + (1 - 1)^2 = 16 \), which confirms the point is on the circle.
Understanding these steps helps in visualizing and accurately drawing circles, making geometry problems easier to tackle.