Problem 45 - Precalculus

Question

Determine the center and the radius for the circle. Also, find the \(y\) -coordinates of the points (if any) where the circle intersects the \(y\) -axis. $$x^{2}+y^{2}+8 x-6 y=-24$$

Step-by-Step Solution

Answer

The center is (-4, 3), the radius is 1, and the circle does not intersect the y-axis.

Step 1: Rearrange the Equation

Start by rearranging the given equation to separate the terms involving \(x\) and \(y\). The equation is:\[x^2 + y^2 + 8x - 6y = -24.\] Group the \(x\) related terms together and the \(y\) related terms together:\[(x^2 + 8x) + (y^2 - 6y) = -24.\]

Step 2: Complete the Square for x

Take the group of terms \(x^2 + 8x\) and complete the square. 1. Take half of the coefficient of \(x\) (which is 8) and square it: \((8/2)^2 = 16\).2. Add and subtract 16 inside the equation:\[(x^2 + 8x + 16 - 16) + (y^2 - 6y) = -24.\]3. The expression \(x^2 + 8x + 16\) becomes \((x + 4)^2\). Replace it in the equation:\[(x + 4)^2 - 16 + (y^2 - 6y) = -24.\]

Step 3: Complete the Square for y

Take the group of terms \(y^2 - 6y\) and complete the square.1. Take half of the coefficient of \(y\) (which is -6) and square it: \((-6/2)^2 = 9\).2. Add and subtract 9 inside the equation:\[(x + 4)^2 - 16 + (y^2 - 6y + 9 - 9) = -24.\]3. The expression \(y^2 - 6y + 9\) becomes \((y - 3)^2\). Replace it in the equation:\[(x + 4)^2 + (y - 3)^2 - 16 - 9 = -24.\]

Step 4: Simplify the Equation

Simplify the equation by moving the constants to the right side:\[(x + 4)^2 + (y - 3)^2 = -24 + 16 + 9.\]Calculate the right side:\[-24 + 16 + 9 = 1.\]Now, the equation is:\[(x + 4)^2 + (y - 3)^2 = 1.\]

Step 5: Determine the Center and Radius

The equation \((x + 4)^2 + (y - 3)^2 = 1\) is now in the standard form of a circle equation:\((x - h)^2 + (y - k)^2 = r^2,\)where \((h, k)\) is the center and \(r\) is the radius of the circle.Thus, the center is \((-4, 3)\) and the radius is \(\sqrt{1} = 1\).

Step 6: Find the Intersection with y-Axis

To find intersection points with the \(y\)-axis, set \(x = 0\) in the equation:\[(0 + 4)^2 + (y - 3)^2 = 1.\]This simplifies to:\[16 + (y - 3)^2 = 1.\]Rearrange and solve for \(y\):\[(y - 3)^2 = 1 - 16.\]\[(y - 3)^2 = -15.\]Since the square of a real number cannot be negative, there are no real \(y\) values that satisfy this.Hence, the circle does not intersect the \(y\)-axis.

Key Concepts

Completing the Square

Completing the square is a technique used to transform quadratic equations into a format that is easier to work with. This method plays a crucial role in many areas, including solving equations of circles as seen in the given problem. In the context of the equation for a circle, completing the square helps in converting the general form of a quadratic equation into the standard form of a circle equation.

The general form of a circle equation is given by the sum of squares of the variables, plus additional linear terms, such as:

• \((x^2 + y^2 + Dx + Ey = F)\)

To complete the square for each variable:

• Take the linear terms, e.g., \(Dx=8x\), and divide the coefficient of the variable by two, then square it.
• Add and subtract this squared value within the equation.
• Reformat the equation into perfect square trinomials.
• For \(x^2 + 8x\), this results in \( (x + 4)^2\).

Similarly, complete the square for the \(y\) terms. Once all terms are completed, the equation can be refactored into the standard form, \((x - h)^2 + (y - k)^2 = r^2\). This form makes it easier to identify the circle's center and radius.

Circle Intersection

Understanding how to find the intersection points of a circle with an axis or another line is vital in geometry. In this exercise, we're focused on the intersection with the \(y\)-axis. When seeking intersection points where a circle meets the \(y\)-axis, we substitute \(x = 0\) in the circle's equation. This is because any point on the \(y\)-axis has an \(x\)-coordinate of 0.

For example, in the equation \((x + 4)^2 + (y - 3)^2 = 1\), substituting \(x = 0\) transforms the equation into:

• \(16 + (y - 3)^2 = 1\)

However, solving leads to \((y - 3)^2 = -15\), which implies no real solutions exist. This conclusion means the circle does not intersect the \(y\)-axis at any point. This highlights that if simplifying reveals a negative on the right side (of the \( (y - 3)^2 = ext{value} \) equation), there cannot be real solutions—emphasizing a key aspect of circle intersections.

Center and Radius of a Circle

Finding the center and radius of a circle from its equation gives us valuable geometric information. Transforming the equation into the standard form \((x - h)^2 + (y - k)^2 = r^2\) reveals these properties clearly.

Using the result of completing the square, the given equation becomes \((x + 4)^2 + (y - 3)^2 = 1\). This is directly comparable to the standard form, where:

• \(h = -4\), indicating the center's \((x)\)-coordinate.
• \(k = 3\), the \((y)\)-coordinate of the center.
• The radius \(r\) is the square root of the right-hand side: \(r = \sqrt{1} = 1\).

Thus, the center of this circle is at \((-4, 3)\) and its radius is \1\. Understanding these transformations from algebra to geometry is crucial, as it allows us to visually interpret and place the circle accurately on a coordinate plane.