Problem 75 - College Algebra

Question

Write the standard form and the general form of the equation of each circle. Passing through \((-7,2)\) and \((1,2) ;\) these points are endpoints of the diameter, the line that passes through the circle's center.

Step-by-Step Solution

Answer

The standard form of the circle's equation is \((x+3)^2 + (y-2)^2 = 16\), and the general form is \(x^2 + y^2 + 6x - 4y + 5 = 0\)

Step 1: Calculate the Circle's Center

Start by finding the circle's center. The center can be found by calculating the mid-point of the diameter. The midpoint formula is \( \frac{{x1+x2}}{2} , \frac{{y1+y2}}{2}\), where \((x1, y1)\) and \((x2, y2)\) are the endpoints of the diameter. For given points, the midpoint is \((-3, 2)\)

Step 2: Calculate the Circle's Radius

Next, calculate the distance from the center of the circle to any of the endpoints of the diameter, which is the radius. The distance formula is \(\sqrt{{(x2-x1)^2 + (y2-y1)^2}}\). Here \((x1, y1)\) is the center and \((x2, y2)\) is one of the endpoints. The radius turns out to be 4.

Step 3: Write the Standard Form of the Circle's Equation

The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Substituting \((-3, 2)\) for \((h, k)\) and 4 for \(r\) gives \((x+3)^2 + (y-2)^2 = 16\)

Step 4: Convert the Equation to General Form

The general form of a circle's equation is \(x^2 + y^2 + Dx + Ey + F = 0\). By expanding and rearranging the standard form equation, we get \(x^2 + y^2 + 6x - 4y + 5 = 0\)

Key Concepts

Standard Form

In circle equations, the standard form is a compact and clean way to express the equation of a circle. The general formula is \[(x-h)^2 + (y-k)^2 = r^2\]where:- \((h, k)\) represents the coordinates of the circle's center- \(r\) is the radius of the circleThis form is quite helpful because it directly shows the center of the circle and its radius.
To find the equation in standard form, we need the circle's center and radius. In the given problem, the center's coordinates are found using the midpoint formula—which averages the coordinates of the two endpoints of the diameter. With the midpoint at \((-3, 2)\) and a radius of 4, the standard form of this circle's equation becomes \[(x+3)^2 + (y-2)^2 = 16\].
Whenever you have a circle's center and radius, fitting them into this formula gives a clear and precise standard form.

General Form

The general form of a circle's equation rearranges the standard form into a polynomial expression. Its general equation looks like: \[x^2 + y^2 + Dx + Ey + F = 0\]You start with the standard form and expand it by squaring the terms and rearranging them to fit this layout. This converts the elegant and direct standard form into a more algebraically complex form.
For our exercise's circle, we expanded the standard form equation \((x+3)^2 + (y-2)^2 = 16\) and, step-by-step, simplified it to become \[x^2 + y^2 + 6x - 4y + 5 = 0\].
While the general form might seem less intuitive at first, it is used in various mathematical applications where quadratic terms need to be compared or integrated into larger systems of equations.

Midpoint Formula

The midpoint formula is an essential tool for finding the center of a circle when given two endpoints of a diameter.
Given two points \((x_1, y_1)\) and \((x_2, y_2)\), the formula for the midpoint is:\[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]This formula takes the average of the x-coordinates and the y-coordinates of the points.
For our specific problem, the diameter's endpoints are \((-7, 2)\) and \((1, 2)\). After applying the midpoint formula, the circle’s center is calculated as \((-3, 2)\).
This midpoint formula provides a straightforward way to determine the center of a circle, which is a critical step before finding the equation in standard form.

Distance Formula

The distance formula helps us calculate the radius of a circle when we know the center point and an endpoint of the diameter. It is given by:\[\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]This formula is derived from the Pythagorean theorem, measuring the direct distance between two points in the Cartesian plane.
In the problem, to find the radius of the circle, we use the center at \((-3, 2)\) and one endpoint \((1, 2)\). Substituting these values, the radius turned out to be 4.
The distance formula is crucial whenever you need to determine how far apart two points are. In circle equations, it helps ensure that you have the correct radius, allowing accurate representation in both the standard and general forms.