Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 47-56: Find the center and radius of the circle with the given equation. $$ x^{2}+y^{2}-4 x+6 y-36=0 $$

Step-by-Step Solution

Answer

The center is (2, -3) and the radius is 7.

Step 1: Write the General Equation of a Circle

The general equation of a circle is given by \[(x - h)^2 + (y - k)^2 = r^2\]where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Rearrange Terms to Complete the Square

The given equation is \[x^2 + y^2 - 4x + 6y - 36 = 0.\]We need to rearrange and complete the square for both \(x\) and \(y\) terms. First, rearrange it to focus on terms involving \(x\) and \(y\):\[x^2 - 4x + y^2 + 6y = 36.\]

Step 3: Complete the Square for x

For the \(x\) terms, \(x^2 - 4x\):1. Take half of \(-4\), which is \(-2\), and square it, resulting in \(4\).2. Add and subtract \(4\) inside the equation:\[x^2 - 4x = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4.\]

Step 4: Complete the Square for y

For the \(y\) terms, \(y^2 + 6y\):1. Take half of \(6\), which is \(3\), and square it, resulting in \(9\).2. Add and subtract \(9\) inside the equation:\[y^2 + 6y = (y^2 + 6y + 9) - 9 = (y + 3)^2 - 9.\]

Step 5: Form the Circle Equation

Substituting the completed squares back into the rearranged equation:\[(x - 2)^2 - 4 + (y + 3)^2 - 9 = 36.\]Combine and balance the constants:\[(x - 2)^2 + (y + 3)^2 = 36 + 4 + 9 = 49.\]

Step 6: Identify the Center and Radius

From the equation \((x - 2)^2 + (y + 3)^2 = 49\), we identify:- The center of the circle is \((h, k) = (2, -3)\).- The radius \(r\) is the square root of 49, which is \(7\).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial, making it easier to work with. For circles, this technique helps us rewrite the equation in a recognizable form to find important geometric properties.
Let's see how this process works step by step for an equation in the form of \(x^2 + bx\):

• Identify the coefficient of the linear term, in this case it’s \(-4\) for \(x\) and \(+6\) for \(y\).
• Divide this coefficient by 2 (resulting in \(-2\) and \(3\), respectively).
• Square the result (giving \(4\) and \(9\), respectively).

Add and subtract these squares within the equation. In our example, adding \(4\) and \(9\) completes the squares for \(x^2 - 4x\) and \(y^2 + 6y\). As a result, the right-hand side remains balanced, allowing the equation to be neatly rewritten as \((x - 2)^2 + (y + 3)^2 = r^2\).
This new equation reveals more about the circle's nature.

Center of a Circle

The center of a circle is a point from which every point on the circle is equidistant. When dealing with equations, the center is represented as coordinates in the form \( (h, k) \).
Once we have transformed the circle's equation using completing the square, we compare it to the standard form \((x - h)^2 + (y - k)^2 = r^2\) to identify the center. For our given equation, it becomes clear:

• The \((x - 2)\) term indicates that the x-coordinate of the center is \2\.
• The \((y + 3)\) term (which can be rewritten as \((y - (-3))\)) tells us that the y-coordinate of the center is \-3\.

Thus, the center of this circle is at the point \( (2, -3) \), providing a fixed point about which the circle is perfectly balanced.

Radius of a Circle

The radius of a circle is the distance from its center to any point on the circle itself, a foundational characteristic of circles. In circle equations, the radius can be determined from the value on the right side of the equation's standard form \((x - h)^2 + (y - k)^2 = r^2\).
In our example, after completing the square, we end up with:

• The equation \((x - 2)^2 + (y + 3)^2 = 49\).
• This tells us \((49 is the squared radius)\).

To find the actual radius \((r)\), we simply take the square root of \(49\), resulting in \(7\).
This calculation shows that every point on the circle is exactly \(7\) units away from the center at \((2, -3)\), giving the circle its size and boundaries.