Question

Find the center and radius of the circle $$ x^{2}+4 x+y^{2}-2 y=11 $$

Short answer

Center: (-2, 1); Radius: 4.

Step-by-step solution

Identify the General Form of the Circle Equation

The given equation of the circle is in the form of \[ x^2 + 4x + y^2 - 2y = 11 \] which can be rewritten to clearly see the terms involving x and y.

Group x and y Terms Together

Rewrite the equation grouping the x and y terms together: \[ (x^2 + 4x) + (y^2 - 2y) = 11 \]

Complete the Square for x

To complete the square for the x terms, take half of the coefficient of x, square it, and add it inside the parenthesis: \[ x^2 + 4x = (x^2 + 4x + 4) - 4 = (x + 2)^2 - 4 \]. Add and subtract 4 inside the equation.

Complete the Square for y

To complete the square for the y terms, take half of the coefficient of y, square it, and add it inside the parenthesis: \[y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1 \]. Add and subtract 1 inside the equation.

Rewrite the Equation

Substitute the completed squares back into the equation: \[ (x+2)^2 - 4 + (y-1)^2 - 1 = 11 \]. Combine the constants on the right side: \[ (x+2)^2 + (y-1)^2 - 5 = 11 \]. Finally, add 5 to both sides to isolate the circle terms: \[ (x+2)^2 + (y-1)^2 = 16 \].

Identify the Center and Radius

The equation is now in the standard form of a circle: \[ (x-h)^2 + (y-k)^2 = r^2 \], where \( h = -2 \), \( k = 1 \), and \( r^2 = 16 \). The center of the circle is \( (-2, 1) \) and the radius is \( r = \sqrt{16} = 4 \).

Key concepts

completing the square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial.
This technique is particularly useful when working with circle equations.
To complete the square, follow these steps:

First, take the coefficient of the linear term (the term with the first degree variable, like x or y).
In our example, the x term is 4x, so the coefficient is 4.
Divide it by 2, then square it: \( \frac{4}{2} = 2 \) and \( 2^2 = 4 \).
Next, add and subtract that square inside the equation to form a perfect square trinomial.
The same process is applied to the y term.
For \( y^2 - 2y \), take \( -2 \), divide by 2 to get \( -1 \), and then square it to get 1.
Completing the square helps easily convert the general form of a circle's equation to its standard form.

center and radius

The center and radius are essential properties of a circle that define its size and location on a coordinate plane.
To find these, we use the standard form of a circle's equation.
For the equation \( (x+2)^2 + (y-1)^2 = 16 \), we can directly identify the center \( (h, k) \) and radius (r).
The general format of the circle equation \( (x-h)^2 + (y-k)^2 = r^2 \) provides a clear way to extract these values:

• \footnotesize The value of h is the x-coordinate of the circle's center.
• \footnotesize The value of k is the y-coordinate of the circle's center.
• \footnotesize The value of r is the circle's radius, found by taking the square root of the constant term on the right side of the equation.


Thus, for our example, the center of the circle is found directly as \( (-2, 1) \) and the radius as \( \sqrt{16} = 4 \).
Understanding these parts will greatly assist in graphing and analyzing circles.

standard form of a circle

The standard form of a circle's equation is crucial for easily identifying the circle's center and radius.
This form is written as \( (x-h)^2 + (y-k)^2 = r^2 \), where (h, k) is the center and r is the radius.
Let's break it down:

• \( x \) and \( y \) are the variables representing coordinates on the plane.
• \( h \) and \( k \) are constants representing the x and y coordinates of the circle's center.
• \( r^2 \) is the squared value of the circle's radius.

The steps to convert a general circle equation to standard form involve grouping the x and y terms and completing the square for each.
By rewriting the generic form into this structured form, we clearly see the circle's key features.
Mastering this conversion is fundamental in understanding how circles operate geometrically.