Problem 13 - Calculus Single Variable

Question

The Cartesian equation of a circle is given. Sketch the circle and specify its center and radius. \(3 x^{2}-6 y+12 x+3 y^{2}=2\)

Step-by-Step Solution

Answer

The center is (-2, 1) and the radius is \(\sqrt{\frac{17}{3}}\).

Step 1: Rearrange Circle Equation

The given equation is \(3x^2 - 6y + 12x + 3y^2 = 2\). First, divide the whole equation by 3 to simplify it: \(x^2 + y^2 + 4x - 2y = \frac{2}{3}\).

Step 2: Group and Complete the Square

Rewrite the equation to group the variables: \((x^2 + 4x) + (y^2 - 2y) = \frac{2}{3}\). Next, complete the square for each group. For \(x^2 + 4x\), add and subtract 4 (\((x^2 + 4x + 4) - 4\)). For \(y^2 - 2y\), add and subtract 1 (\((y^2 - 2y + 1) - 1\)).

Step 3: Simplify the Equation

Rewrite the equation using the completed squares: \((x + 2)^2 - 4 + (y - 1)^2 - 1 = \frac{2}{3}\). Simplify to: \((x + 2)^2 + (y - 1)^2 = \frac{2}{3} + 5\).

Step 4: Calculate Simplified Constant

Calculate the simplified constant on the right: \((x + 2)^2 + (y - 1)^2 = \frac{17}{3}\). This equation is in the standard form of a circle equation, \((x - h)^2 + (y - k)^2 = r^2\).

Step 5: Identify Center and Radius

From the equation \((x + 2)^2 + (y - 1)^2 = \frac{17}{3}\), identify the center \((-2, 1)\) and calculate the radius \(r = \sqrt{\frac{17}{3}}\).

Key Concepts

Center of Circle

When we talk about the center of a circle in the context of its Cartesian equation, we're referring to the point in the Cartesian plane from which all points on the circle are equidistant. Consider the standard form of the circle's equation:
\((x - h)^2 + (y - k)^2 = r^2\). Here, the center of the circle is

• h: the x-coordinate of the center
• k: the y-coordinate of the center

In our case, from the provided circle equation
\((x + 2)^2 + (y - 1)^2 = \frac{17}{3}\), we can extract the center by comparing it to the standard form.Observe:

• The expression
  \(x + 2\) can be rewritten as \((x - (-2))\), indicating the center's x-coordinate is \(-2\).
• The expression
  \(y - 1\) remains unchanged, so the center's y-coordinate is \(1\).

Thus, the center of this circle is
\((-2, 1)\). This means if you were to plot this circle on a graph, you would start at point
\((-2, 1)\) as the focal point of the circle.

Radius of Circle

The radius of a circle is the distance from its center to any point on its outer edge. In the equation
\((x - h)^2 + (y - k)^2 = r^2\), the term \(r^2\) is a critical part of understanding a circle’s radius.In the simplified equation from our example
\((x + 2)^2 + (y - 1)^2 = \frac{17}{3}\), the right side of the equation is set equal to
\(r^2\). Thus, the radius squared (\(r^2\)) is
\(\frac{17}{3}\).To find the radius \(r\),
we take the square root of \(\frac{17}{3}\):

• Calculating the Square Root:
  \(r = \sqrt{\frac{17}{3}}\)

This calculation provides the length of the radius. In practical terms, it means every point on the circle's edge is exactly
\(\sqrt{\frac{17}{3}}\) units away from its center,
\((-2, 1)\).

Completing the Square

Completing the square is a fundamental algebraic technique used to transform a quadratic expression into a perfect square trinomial.For the equation of a circle, it's often necessary to convert terms like
\(x^2 + bx\) into a square form to easily identify the circle's properties, like its center and radius.In our task, we started with
\(x^2 + 4x\) and \(y^2 - 2y\):

• For \(x^2 + 4x\):
  Add and subtract 4 to complete the square:
  \(x^2 + 4x + 4 - 4\).
  This results in the perfect square
  \((x + 2)^2 - 4\).
• For \(y^2 - 2y\):
  Add and subtract 1:
  \(y^2 - 2y + 1 - 1\).
  This results in the perfect square
  \((y - 1)^2 - 1\).

The process of completing the square rearranges the quadratic expression into a form that highlights the center of the circle as
\((-2, 1)\) and converts it into an easily interpretable format, as seen in the final equation form
\((x + 2)^2 + (y - 1)^2 = \frac{17}{3}\). This highlights both the geometry and the algebra involved.