Problem 60 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 57-60: Find equations for the upper half, lower half, right half, and left half of the circle. $$ (x-3)^{2}+(y-5)^{2}=4 $$

Step-by-Step Solution

Answer

Upper half: \( y = 5 + \sqrt{4 - (x-3)^2} \), Lower half: \( y = 5 - \sqrt{4 - (x-3)^2} \), Right half: \( x = 3 + \sqrt{4 - (y-5)^2} \), Left half: \( x = 3 - \sqrt{4 - (y-5)^2} \).

Step 1: Understand the Circle Equation

The given equation \((x-3)^{2} + (y-5)^{2} = 4\) represents a circle with center at \((3,5)\) and radius \(2\) (since \(4 = 2^2\)).

Step 2: Recall How to Split the Circle

To find equations representing halves of a circle, we can manipulate the original equation to isolate the parts for the upper, lower, left, and right halves.

Step 3: Equation for the Upper Half

For the upper half, solve for \(y\) in terms of \(x\):\[ y - 5 = \, + \, \sqrt{4 - (x-3)^2} \]Thus, the equation is:\[ y = 5 + \, \sqrt{4 - (x-3)^2} \]

Step 4: Equation for the Lower Half

For the lower half, solve for \(y\) in terms of \(x\):\[ y - 5 = \, - \, \sqrt{4 - (x-3)^2} \]Thus, the equation is:\[ y = 5 - \, \sqrt{4 - (x-3)^2} \]

Step 5: Equation for the Right Half

For the right half, solve for \(x\) in terms of \(y\) where \(x\) is greater than or equal to the center \(x\)-coordinate:\[ x - 3 = \, + \, \sqrt{4 - (y-5)^2} \]Thus, the equation is:\[ x = 3 + \, \sqrt{4 - (y-5)^2} \]

Step 6: Equation for the Left Half

For the left half, solve for \(x\) in terms of \(y\) where \(x\) is less than or equal to the center \(x\)-coordinate:\[ x - 3 = \, - \, \sqrt{4 - (y-5)^2} \]Thus, the equation is:\[ x = 3 - \, \sqrt{4 - (y-5)^2} \]

Key Concepts

Circle Properties

Circles are fascinating geometric shapes with unique properties that make them distinct from other shapes. A circle is defined as the set of all points in a plane that are equidistant from a given point, called the center. This constant distance is known as the radius. In the circle equation

• he standard form is: \((x - h)^2 + (y - k)^2 = r^2\)
• where \(h, k\) represents the center while \(r\) is the radius.

In our exercise, the given equation \((x-3)^2 + (y-5)^2 = 4\) identifies a circle with

• a center at \((3,5)\)
• and a radius of \(2\).

This is because \(4 = 2^2\).The radius determines the size of the circle, while the center controls its position in the coordinate plane.

Equation Solving

Understanding how to work with circle equations involves solving them to find specific parts or separate components. We can split a circle into different halves: upper, lower, left, and right.To solve for these halves, we manipulate the original circle equation:

• The upper and lower halves involve isolating \(y\) in terms of \(x\), using square roots to find the points above or below the circle's center line.
• The right and left halves involve isolating \(x\) in terms of \(y\), focusing on the points to the right or left of the center.

This requires careful algebraic steps: taking square roots, considering positive or negative results, and ensuring you maintain the radius as defined in the original equation. Solving the circle equation like this allows us to examine smaller sections selectively.

Geometric Transformations

Geometric transformations refer to the changes in position, size, or shape of a geometric figure. Circles, when transformed, can maintain their original shape while moving across the coordinate plane. Working with our circle equation \((x-3)^2 + (y-5)^2 = 4\), we see no rotational or size transformation, but the position of the circle has shifted due to its center:

• Translating by changing the center's coordinates from \((0, 0)\) to \((3, 5)\).
• Such translation results in the circle maintaining its original size and shape, just relocated on the plane.

This aspect of transformation highlights the power of **coordinate forms** to dictate geometric forms across different positions.

Coordinate Geometry

Coordinate geometry, or analytic geometry, allows us to study geometric shapes strictly by employing algebraic expressions in the coordinate plane. With circles, we derive equations like \((x-3)^2 + (y-5)^2 = 4\) to make precise calculations on graph-paper-like frameworks. Key benefits include:

• Determining the exact distance between points with formulas derived from the Pythagorean theorem.
• Finding intersections, midpoints, and symmetry by simple algebraic manipulation of equations.
• Solving more complex problems involving the curvature of arcs or tangents requires just palms on equations, leading to solid geometry contributions.

Thus, **coordinate geometry** is a robust tool supporting depth studies into circles and other curves, allowing for detailed explorations outside simple sketches.