Problem 21 - Algebra 2

Question

For each equation, find the center and radius of the circle. $$ (x-3)^{2}+(y+1)^{2}=36 $$

Step-by-Step Solution

Answer

The center of the circle is at \((3,-1)\) and the radius is 6.

Step 1: Identify the Center

The center \((h,k)\) of the circle is determined by the coefficients of \(x\) and \(y\) in the equation. From the equation \((x-3)^{2}+(y+1)^{2} =36 \), we can conclude that \(h=3\) and \(k=-1\). Thus, the center of the circle is \((3,-1)\).

Step 2: Identify the Radius

The radius \(r\) of the circle is determined by the constant on the right side of the equation. The given equation is of the form \((x-h)^{2}+(y-k)^{2} = r^{2}\), and in our equation, \(r^{2} =36\). So, to find \(r\), we take the square root of 36, i.e, \(r=6\). Thus, the radius of the circle is 6.

Key Concepts

Center of Circle

Understanding the center of a circle is crucial in analyzing its equation. For the given equation, \((x-3)^{2} + (y+1)^{2} = 36\), the center is derived from the terms \((x-h)^{2}\) and \((y-k)^{2}\). Here,

• \(h\) represents the horizontal shift of the circle's center from the origin.
• \(k\) stands for the vertical shift.

To find these values, we compare our equation to the standard form \((x-h)^{2} + (y-k)^{2} = r^{2}\). In this case, where \(h = 3\) and \(k = -1\), the center is at point \((3, -1)\). Knowing the center helps visualize the circle's location on the coordinate plane. Breaking down the shifts: - The \(x-h\) term shows a move 3 units to the right.- The \(y-k\) term moves it 1 unit down. Altogether, this places our circle's center clearly at \((3, -1)\).

Radius of Circle

The radius of a circle is the distance from the center to any point along its edge. It is a vital part of understanding circle geometry. In the problem's equation, \((x-3)^{2} + (y+1)^{2} = 36\), once you have identified the center, the next step is to find the **radius**.In the standard equation form \((x-h)^{2} + (y-k)^{2} = r^{2}\), \(r^{2}\) is represented by the constant on the right. Here, it is 36. To find the actual radius, you take the square root of \(r^{2}\):

• \(r = \sqrt{36}\)
• \(r = 6\)

This calculation concludes that the circle has a radius of 6. This distance from its center affects its size, as it indicates how far out the circle will extend in all directions from its center.

Standard Form of Circle Equation

The standard form of a circle's equation is the starting point for any analysis involving circles. In the format \((x-h)^{2} + (y-k)^{2} = r^{2}\):

• \(h\) and \(k\) define the circle's center.
• \(r\) represents the radius.

This form allows you to quickly determine the circle's key features from the equation itself. By recognizing the standard form, you can easily identify both the center and radius, which are essential for graphing and understanding the circle's geometry.In our given problem \((x-3)^{2} + (y+1)^{2} = 36\), the form remains consistent with the standard, letting us decompose it directly:- \(x-3\) tells us the horizontal center shift.- \(y+1\) indicates the vertical center shift.By identifying these components, you unlock the ability to solve for the circle's properties, making this form vital for students dealing with circle equations.