Problem 47 - 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. $$9 x^{2}+54 x+9 y^{2}-6 y+64=0$$

Step-by-Step Solution

Answer

The center is \((-3, \frac{1}{3})\), radius is \(\sqrt{2}\), and the circle does not intersect the \(y\)-axis.

Step 1: Write the Standard Equation of a Circle

The standard equation of a circle is given by \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius. We need to manipulate the given equation to this form.

Step 2: Rearrange the Equation

Rewrite the given equation by grouping \(x\) and \(y\) terms:\[ 9(x^2 + 6x) + 9(y^2 - \frac{2}{3}y) = -64 \]

Step 3: Complete the Square for x-terms

Complete the square for \(x^2 + 6x\):\[ x^2 + 6x = (x+3)^2 - 9 \] Substitute back to get:\[ 9((x+3)^2 - 9) = 9(x+3)^2 - 81 \]

Step 4: Complete the Square for y-terms

Complete the square for \(y^2 - \frac{2}{3}y\):\[ y^2 - \frac{2}{3}y = \left(y - \frac{1}{3}\right)^2 - \frac{1}{9} \]Substitute back to get:\[ 9\left(\left(y - \frac{1}{3}\right)^2 - \frac{1}{9}\right) = 9\left(y - \frac{1}{3}\right)^2 - 1 \]

Step 5: Simplify and Rearrange the Equation

Combine all parts:\[ 9(x+3)^2 - 81 + 9\left(y - \frac{1}{3}\right)^2 - 1 = -64 \] Simplify:\[ 9(x+3)^2 + 9\left(y - \frac{1}{3}\right)^2 = 18 \] Divide by 9:\[ (x+3)^2 + \left(y - \frac{1}{3}\right)^2 = 2 \]

Step 6: Identify the Center and the Radius

From the equation \((x+3)^2 + \left(y - \frac{1}{3}\right)^2 = 2 \), the center is \((-3, \frac{1}{3})\) and the radius is \(\sqrt{2}\).

Step 7: Find Points Where the Circle Intersects the y-axis

To find intersection points with the \(y\)-axis, set \(x = 0\) in the circle equation:\[ (0+3)^2 + \left(y - \frac{1}{3}\right)^2 = 2 \]Simplify:\[ 9 + \left(y - \frac{1}{3}\right)^2 = 2 \]\[ \left(y - \frac{1}{3}\right)^2 = -7 \]Since \(-7\) is negative, there are no real intersection points with the \(y\)-axis.

Key Concepts

Completing the Square

Understanding how to complete the square is key when working with quadratic expressions, especially in the context of circle equations. Imagine you have an expression like \( x^2 + 6x \). This is not a perfect square trinomial, but we can turn it into one by completing the square.

The goal is to form something that looks like \( (x + p)^2 \), which is a perfect square trinomial. Here's what you do:

• Take the coefficient of the linear \(x\) term (which is 6), divide it by 2, and square it: \((6/2)^2 = 9\).
• Add and subtract this number within the same expression to balance it out.

This turns \( x^2 + 6x \) into \( (x + 3)^2 - 9 \). You essentially figure out what makes the quadratic into a perfect square and then adjust for the constant term outside the bracket.

Completing the square is also applied to the \( y \) terms in the same way, ensuring you can rewrite the entire circle's equation in a more recognizable form.

Standard Form of a Circle

The standard form of a circle equation helps identify crucial features about the circle effortlessly. Consider the form: \( (x-h)^2 + (y-k)^2 = r^2 \).

Here, \((h, k)\) represents the center of the circle and \(r\) is the radius. Once we've rewritten a given equation into standard form, finding the center and the radius is straightforward.

For example, from the equation: \( (x+3)^2 + (y - \frac{1}{3})^2 = 2 \). The negative of \(+3\) and \(-\frac{1}{3}\) immediately give you the center \((-3, \frac{1}{3})\). The right side of the equation, \(2\), identifies \(r^2\), so the radius \(r\) is \(\sqrt{2}\).

Recasting equations into this standard form not only reveals geometric properties but also simplifies solving problems tied to geometry and intersection.

Intersecting the Y-axis

Determining where a circle crosses the \(y\)-axis involves some simple substitutions and calculations. To find \(y\)-axis intersections, set \(x = 0\) in the circle's equation because any point on the \(y\)-axis will have \(x\)-coordinate equal to zero.

From the given example, substitute \(x = 0\) into: \( (0+3)^2 + (y - \frac{1}{3})^2 = 2 \).

Simplify this to get \(9 + (y - \frac{1}{3})^2 = 2\).

Noticeably, this simplifies to: \( (y - \frac{1}{3})^2 = -7 \). Since the square of a real number can't be negative, this tells us there are no real values for \(y\) that satisfy this when \(x = 0\). Hence, the circle does not intersect the \(y\)-axis.

Through this method, we can always verify if and where a circle crosses the \(y\)-axis by checking if the equation holds true for real values.