Problem 34 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 23-34: Sketch the graph of the circle or semicircle. $$ x=-\sqrt{25-y^{2}} $$

Step-by-Step Solution

Answer

The semicircle is on the left (negative x) side of the y-axis, centered at the origin, with radius 5.

Step 1: Recognize the Equation Type

The given equation is in the form \(x = -\sqrt{25-y^2}\), which is a version of a circle equation. Normally, the circle equation is \(x^2 + y^2 = r^2\). Here, \(x = -\sqrt{25-y^2}\) suggests a semicircle.

Step 2: Identify the Radius

In the equation \(x = -\sqrt{25-y^2}\), the expression under the square root, \(25-y^2\), implies that the radius \(r\) of the circle is 5, because \(25 = r^2\).

Step 3: Determine the Circle's Center and Direction

Since the equation is \(x = -\sqrt{25-y^2}\), it indicates a circle centered at the origin \((0, 0)\) but only on the left side of the y-axis, due to the negative square root, forming a semicircle.

Step 4: Plot the Semicircle

To draw the semicircle, plot points starting from the top \((0, 5)\) to the bottom \((0, -5)\) of the circle on the y-axis, ensuring all x-values are negative. Since \(x = -\sqrt{25-y^2}\), compute x for various y-values: \(x = -5\) at \(y=0\), \(x = 0\) at \(y=5\) and \(y=-5\). Connect these smoothly.

Step 5: Verify the Sketch

Ensure that the plotted points correctly fit the equation \(x = -\sqrt{25-y^2}\) by substituting back: For instance, if \(y=3\), \(x = -\sqrt{25 - 3^2} = -\sqrt{16} = -4\). The semicircle should appear on the left half of the \(y-axis\), from \((0,5)\) to \((0,-5)\).

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of geometry where algebraic equations are used to describe geometric figures. It combines algebra and geometry by utilizing the coordinate planes (x-axis and y-axis) to visualize shapes and solve problems involving their properties. In coordinate geometry, each point is identified by an x-coordinate and a y-coordinate. This allows us to graph lines, curves, and other shapes with precision.
For circles and semicircles, coordinates help us determine positions of points on these curves, allowing for accurate plotting and analysis. For example, using coordinates, one can easily find the intersection points of a line and a circle, or describe the movement from the top to the bottom of a semicircle.
With tools such as the distance formula and midpoint formula, coordinate geometry proves valuable in solving various real-world and theoretical problems.

Semicircle

A semicircle is exactly half of a circle. It is formed when a circle is divided along its diameter. The properties of a semicircle are similar to those of a full circle but applied to only one half.
The diameter of the semicircle becomes its base, while the curved edge is called the arc. In coordinate geometry, a semicircle can be expressed with equations such as the given one: \(x = -\sqrt{25-y^2}\). This indicates that only the left half of the circle, lying on the negative x-axis, is considered.
Key features of a semicircle include:

• One flat side, which is the diameter.
• The curved arc, which spans 180 degrees or \(\pi\) radians.
• Passage through points of symmetry, reflecting across the diameter.

Understanding the properties of a semicircle is vital for accurate graphing and for solving geometric problems involving semicircular shapes.

Equation of a Circle

The equation of a circle is a mathematical expression that represents all the points that form a circle on the coordinate plane. The standard form of a circle's equation centered at the origin is \(x^2 + y^2 = r^2\), where \(r\) represents the radius.
In the context of the exercise provided, the equation \(x = -\sqrt{25-y^2}\) requires some manipulation to fit the standard form. This version represents a circle (specifically a semicircle) centered at the origin, only on the negative x-axis. The equation describes only half because of the negative square root, showing that all x-values are on the negative side.
To graph a circle from its equation:

• Start by identifying the radius \(r\) from \(r^2\), in this case, \(r^2 = 25\).
• Plot the center, which here is at the origin \((0,0)\).
• Use the circle equation to compute coordinates of points and plot them.

This systematic approach helps visualize the circle or semicircle on the graph.

Radius of a Circle

The radius of a circle is the distance from the center of the circle to any point on its circumference. It is a crucial component in defining and understanding the properties of a circle.
In the equation \(x^2 + y^2 = r^2\), the radius \(r\) can be identified as the square root of the value on the right side of the equation. For the equation in the task, \(25-y^2 = r^2\), making \(r = 5\).
The radius plays a central role in other calculations involving circles:

• It determines the size of the circle or semicircle.
• It influences the arc length, area, and circumference calculations.
• Helps in locating points on the circle during plotting.

Knowing the radius allows us to ensure accurate drawings and solutions in geometry problems involving circles and semicircles.