Problem 76 - College Algebra with Modeling and Visualization

Question

Find the standard equation of a circle that satisfies the conditions. Center \((3,-5)\) with the point \((4,2)\) on the circle

Step-by-Step Solution

Answer

The standard equation is \((x-3)^2 + (y+5)^2 = 50\).

Step 1: Identify Circle's Components

To find the standard equation of a circle, use the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) are the coordinates of the center and \(r\) is the radius. Here, the given center is \((3, -5)\), so \(h = 3\) and \(k = -5\).

Step 2: Calculate the Radius

The radius \(r\) is the distance between the center \((3, -5)\) and the point \((4,2)\) on the circle. Use the distance formula: \[ r = \sqrt{(4-3)^2 + (2- (-5))^2} \]. Calculate the expression under the square root: \((4-3)^2 = 1^2 = 1\) and \((2 - (-5))^2 = 7^2 = 49\). So \( r = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}\).

Step 3: Write the Circle's Equation

Substitute the center \((h, k) = (3, -5)\) and the radius \(r = 5\sqrt{2}\) into the standard circle equation: \[(x-3)^2 + (y+5)^2 = (5\sqrt{2})^2\].Simplify the right side: \[(5\sqrt{2})^2 = 25 \times 2 = 50\].Thus, the equation of the circle is: \[(x-3)^2 + (y+5)^2 = 50\].

Key Concepts

Center and Radius

Every circle in a plane can be uniquely represented using its center and radius. The center of a circle is a point in the plane, usually denoted as \((h, k)\), from which all points on the circle are equidistant. The radius is this constant distance from the center to any point on the circle. Knowing these two components is crucial because they define the circle entirely.

In the given exercise, the center is specified as \((3, -5)\). To understand this, imagine a point at \(x=3\) on the horizontal axis and \(y=-5\) on the vertical axis. This point serves as the circle's center. Next, consider the radius. It determines how large the circle is by specifying the distance from this center to any point on the circle itself. This radius is not given directly in the exercise, so you will need to calculate it using other information provided.

Distance Formula

The distance formula is a fundamental tool in geometry to determine how far apart two points are in the Cartesian coordinate system. This is particularly important for circles where you often need to find the radius when only given two points: the center \((h, k)\) and any point \((x_1, y_1)\) on the circle.

The distance formula is expressed as:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This formula essentially applies the Pythagorean theorem to find the diagonal distance between two points.

In this exercise, the center of the circle is \((3, -5)\) and the point on the circle is \((4, 2)\). Plug in these values into the distance formula:

• \( r = \sqrt{(4-3)^2 + (2 + 5)^2} \)

Calculate each part:

• \((4-3)^2 = 1^2 = 1\)
• \((2 + 5)^2 = 7^2 = 49\)

So, \( r = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \). This calculated value is the radius of the circle.

Standard Form of Circle

The standard form of a circle's equation is a neat, compact way to express all the essential information about a circle on a coordinate plane. It allows you to easily identify the center and calculate the radius.

The standard form is given by:

• \((x-h)^2 + (y-k)^2 = r^2 \)

Here, \((h, k)\) represent the center of the circle, while \(r\) is the radius. Every term in this equation has its role in describing the circle.

In the specific problem, the center is \((3, -5)\) and the radius has been calculated as \(5\sqrt{2}\). Substitute these values:

• \((x-3)^2 + (y+5)^2 = (5\sqrt{2})^2 \)

Simplify the right side using the properties of exponents:

• \((5\sqrt{2})^2=25 \times 2=50\)

Thus, the standard equation of the circle becomes:

• \((x-3)^2 + (y+5)^2 = 50 \)

This equation beautifully captures the circle's geometry in a mere math sentence!