Problem 93 - College Algebra

Question

The graph of \((x-3)^{2}+(y+5)^{2}=-36\) is a circle with radius 6 centered at \((3,-5)\)

Step-by-Step Solution

Answer

The equation does not represent a circle in the standard form since the value corresponding to the radius squared is negative. There seems to be a mistake in the problem statement. If the given equation was \((x-3)^{2}+(y+5)^{2}=36\), then the circle would have center at (3, -5) and radius 6.

Step 1: Identify the form of the standard equation for a circle

Firstly, it's essential to remember that the standard form of the equation for a circle is \((x-a)^2 + (y-b)^2 = r^2\). In this equation, the center of the circle is described by the point \((a, b)\), and r refers to the radius of the circle.

Step 2: Compare the given equation to the standard circle equation

Now, comparing the given equation \((x-3)^{2}+(y+5)^{2}=-36\) with the standard form, we can immediately infer that the values for a and b are 3 and -5, respectively.

Step 3: Identify the radius of the circle

Noticing the right-hand side of the equation, we find it equal to -36. However, the radius squared (\(r^2\)) of a circle cannot be a negative value. This means, there's an inconsistency with the given equation, and it appears to be a mistake. The equation \((x-3)^{2}+(y+5)^{2}=-36\) does not represent a circle in the standard form because the value for \(r^2\) is negative. So, in this context, the radius cannot be found from the given equation.

Key Concepts

Standard Form of a Circle

The standard form of a circle in mathematics is expressed as \((x-a)^2 + (y-b)^2 = r^2\). This formula is key for many students when graphing circles or understanding their geometric properties. Familiarize yourself with this form as it highlights the position and size of a circular graph.

• \( (x-a)^2 + (y-b)^2 \) demonstrates a mathematical equality involving squared binomials, indicating the equation is derivative of the Pythagorean theorem, tailored for circles.
• The variables \(a\), \(b\), and \(r\) in the equation serve special purposes, each representing crucial characteristics of a circle.

The left side, \((x-a)^2 + (y-b)^2\), highlights how the circle's nature is deduced from distances relative to its center. The right side simply gives the radius squared, emphasizing the distance from the center to any point on the boundary of the circle.

Center of a Circle

Located centrally in the standard circle equation are the coordinates \((a, b)\). These numbers tell you where the center of the circle is based on the Cartesian plane. It's always a pair of numbers, denoting a specific point in 2D space.

• Point \((a, b)\) dictates the middle around which the entire shape of the circle revolves.
• From this central point, you can measure equally in all directions to form the boundary, or perimeter, of the circle.

To determine the center of the circle from its equation, compare the expression in the equation to the general form \((x-a)^2 + (y-b)^2\). Here, \(a\) is subtracted from \(x\) and \(b\) is subtracted from \(y\), hence giving the center as \((a, b)\). For the equation \((x-3)^2 + (y+5)^2 = -36\), the center would theoretically be at \((3, -5)\), though remember in this exercise, it carries an issue due to the negative equation component.

Radius of a Circle

Another critical aspect of understanding circles is determining their size, using the radius. The radius is the length from the center of the circle to any point on its circle edge, which is consistent at every direction from the center.

• The part of the circle's equation that reads as \(r^2\) is specifically linked to the circle's size.
• To find the actual length of the radius, you normally take the positive square root of \(r^2\).

When examining \((x-a)^2 + (y-b)^2 = r^2\), \(r^2\) must be a positive number, as it represents a real distance. If the exercise provides a negative result, such as \(-36\), then there's an error or misunderstanding. A correct equation would have a positive right-hand side, ensuring a valid radius could be extracted as \(r = \sqrt{r^2}\). For instance, tweaking the problem to \((x-3)^2 + (y+5)^2 = 36\) would yield \(r = 6\), reflecting a genuine, plausible circle.