Problem 59 - Algebra and Trigonometry

Question

55–62 ? Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are \(P(-1,1)\) and \(Q(5,9)\)

Step-by-Step Solution

Answer

The equation of the circle is \((x - 2)^2 + (y - 5)^2 = 25\).

Step 1: Find the center of the circle (the midpoint of the diameter)

To find the center of the circle, we need to locate the midpoint of the diameter defined by the two endpoints given, points \(P(-1,1)\) and \(Q(5,9)\). The formula for the midpoint \((x, y)\) of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is: \[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]Substitute the coordinates of \(P\) and \(Q\) into the formula:\[\left( \frac{-1 + 5}{2}, \frac{1 + 9}{2} \right) = \left( 2, 5 \right)\]Thus, the center of the circle is \((2, 5)\).

Step 2: Determine the radius of the circle

Next, we calculate the radius of the circle. The radius is half of the diameter, which is the distance between the two points \(P(-1,1)\) and \(Q(5,9)\). Use the distance formula:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substitute the values:\[d = \sqrt{(5 + 1)^2 + (9 - 1)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]The diameter is 10, hence the radius \(r\) is:\[r = \frac{10}{2} = 5\]

Step 3: Write the equation of the circle

With the center \((2, 5)\) and radius \(5\), we can write the equation of the circle using the standard form:\[(x - h)^2 + (y - k)^2 = r^2\]Where \((h, k)\) is the center and \(r\) is the radius. Plugging in our values:\[(x - 2)^2 + (y - 5)^2 = 5^2\]Simplifying:\[(x - 2)^2 + (y - 5)^2 = 25\]This is the equation of the circle.

Key Concepts

Midpoint Formula

The midpoint formula is a powerful tool used in geometry to find the exact center of a line segment. This is especially useful when dealing with problems related to circles, as it allows you to easily determine the circle's center when given the endpoints of a diameter. The formula is expressed as: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]This formula calculates the average of the x-coordinates and the y-coordinates of the two given points. By inputting the coordinates of endpoints P(-1,1) and Q(5,9), we can find the midpoint:

• Calculate the x-coordinate: \( \frac{-1 + 5}{2} = 2 \)
• Calculate the y-coordinate: \( \frac{1 + 9}{2} = 5 \)

So, the center of the circle, which is the midpoint, turns out to be \((2, 5)\). This center point is crucial for establishing the circle's equation and providing a reference for other calculations, such as finding the radius.

Distance Formula

The distance formula derives its utility from the ability to calculate the length between two points in a plane. It is foundational in various fields, including physics and engineering, aside from geometry. The formula is:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]By using this formula, you can determine the distance between points, which, in the context of circle geometry, often refers to finding the diameter or radius.Let's put this into practice for endpoints P(-1,1) and Q(5,9):

• Find the difference in x-coordinates: \(5 - (-1) = 6\)
• Find the difference in y-coordinates: \(9 - 1 = 8\)
• Plug these into the formula: \[d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]

The result shows that the length of the diameter is 10, and the radius of the circle, being half of the diameter, is hence 5. Recognizing and correctly using the distance formula is vital in verifying calculations and ensuring accuracy in creating equations for circles.

Circle Geometry

Circle geometry revolves around understanding the properties and measurements of circles. One of the fundamental aspects is writing the circle's equation. The standard equation of a circle is expressed as:\[(x - h)^2 + (y - k)^2 = r^2\]Here, \((h, k)\) represent the center of the circle, and \(r\) is the radius. Knowing the center, \((2, 5)\), and the radius, 5, allows us to easily plug these values into the equation, resulting in:\[(x - 2)^2 + (y - 5)^2 = 25\]This equation gives a perfect representation of a circle and can be adapted to model real-world problems involving circular paths or areas.Understanding circle geometry will also allow you to work effortlessly with related concepts, such as tangents, chords, and arcs. With the essentials of the standard equation clear, you can apply this knowledge to characterize and solve complex geometric problems concerning circles.