Problem 31 - Thomas Calculus

Question

Graph the lower half of the circle defined by the equation $$x^{2}+2 x=4+4 y-y^{2}$$

Step-by-Step Solution

Answer

Graph the semicircle \(y = 2 - \sqrt{3 - (x+1)^2}\) centered at \((-1, 2)\).

Step 1: Rewrite the equation

First, rearrange the given equation to standard form. Start with \(x^2 + 2x = 4 + 4y - y^2\). Move all terms to one side to get: \(x^2 + 2x - 4 - 4y + y^2 = 0\).

Step 2: Complete the square for the quadratic terms

To rewrite the equation more conveniently, complete the square for both \(x\) and \(y\) terms. For \(x^2 + 2x\), add and subtract \(1\) \((\frac{2}{2})^2\): \((x + 1)^2 - 1\). For \(4 - 4y + y^2\), it's \(-(y - 2)^2 + 4\).

Step 3: Substitute the completed squares into the equation

Substitute back into the equation: \((x+1)^2 - 1 - (y - 2)^2 + 4 = 0\). Simplify to \((x+1)^2 = (y - 2)^2 - 3\).

Step 4: Identify the standard form of the circle

Observe that the equation represents a circle in the form \((x+1)^2 + (y-2)^2 = 3^2/7\). This equation shows a circle centered at \((-1, 2)\) with a radius of \(2\).

Step 5: Determine the lower half of the circle

For the lower half of the circle, solve for \(y\) and choose the negative square root: \(y - 2 = - \sqrt{3 - (x+1)^2}\). Thus, \(y = 2 - \sqrt{3 - (x+1)^2}\). Graph this equation for values of \(x\) such that \(3 - (x+1)^2 \geq 0\).

Step 6: Graph the function

On the Cartesian plane, plot the semicircle centered at \((-1, 2)\) with its radius touching points such that \(x^2 + 2x - 4 = (y - 2)^2\) lying on or below the x-axis.

Key Concepts

Completing the Square

Completing the square is a method used to transform quadratic equations into a form that is easy to analyze and understand. This technique is especially useful when dealing with equations such as circle equations or when solving quadratic functions.

• To complete the square, rearrange terms to separate the \(x\) and \(y\) components.
• For the quadratic expression \(x^2 + 2x\), we add and subtract the square of half of the coefficient of \(x\), which is \((\frac{2}{2})^2 = 1\).
• This results in \((x + 1)^2 - 1\), turning the expression into a perfect square trinomial.

For the \(y\) terms in \(4 - 4y + y^2\), similarly complete the square: \(-(y - 2)^2 + 4\). Now both expressions are in a perfect square form, allowing us to further simplify and solve the circle equation.

Standard Form of Circle Equation

The standard form of a circle equation is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. This form makes it easy to identify the fundamental properties of the circle.

• Upon simplifying our equation through completing the square, it transforms into a format resembling the standard circle equation.
• Our equation becomes $(x+1)^2 + (y-2)^2 = r^2$ , with the center at $(-1, 2)$ .
• The radius $r$ is derived from the equation and is equal to the square root of the constant on the right: hence $r = rac{3^{1/2}}{7}$ , indicating a circle with a known size.

This standardized form allows for clear graphing and analysis of how the circle behaves within a coordinate plane.

Coordinate Geometry

Coordinate Geometry, or analytic geometry, is the study and representation of geometric figures using a coordinate plane. This approach is instrumental for graphing circles and semicircles, as it allows us to express geometrical figures algebraically.

• Using a Cartesian coordinate system, each point on the plane is defined by an ordered pair $(x, y)$ .
• Analyzing circle equations like $(x+1)^2 + (y-2)^2 = rac{3}{7}$ involves identifying all points $(x, y)$ that satisfy the equation within the plane.
• Coordinate geometry enables us to visualize the positioning and dimensions of the circle or semicircle by plotting relevant points, using key elements like the center and radius to determine boundaries and intersections.

This method is a powerful tool for resolving geometrical problems and aids in clear and accurate depiction on graphs.

Semicircle Graphing

Graphing a semicircle involves plotting half of a circle on a coordinate plane. Often, this requires solving the circle's equation for \(y\), then identifying the region corresponding to either the top or bottom half.

• For the lower half of a circle, solve for \(y\) using the equation \(y - 2 = - \sqrt{3 - (x+1)^2}\).
• These points create the curve of the semicircle below the center point \((-1, 2)\) on the graph.
• To properly graph, ensure \(3 - (x+1)^2 \geq 0\), guaranteeing \(y\) is real and the values fall within the permissible domain.

Understanding semicircle graphing helps to distinguish between full and partial representations of circles and efficiently depict specific, defined arcs on a coordinate graph.