Problem 17
Question
Identify the center and radius of the circle. $$(x+1)^{2}+(y+6)^{2}=19$$
Step-by-Step Solution
Verified Answer
The center of the circle is at (-1,-6) and the radius is \(\sqrt{19}\)
1Step 1: Identify the Center of the Circle
In the standard form of a circle equation, the center (h,k) is represented by the numbers that make the expressions inside the parentheses equals zero. Our equation is \((x+1)^{2}+(y+6)^{2}=19\), thus by setting \(x+1 = 0\) and \(y+6 = 0\), it gives \(x = -1\) and \(y = -6\). Therefore, the center (h,k) of the circle is (-1,-6)
2Step 2: Identify the Radius
The radius is given by the square root of the number on the right side of the equation, it's \(\sqrt{r^{2}}\). Thus, the radius \(r\) of the circle is \(\sqrt{19}\)
Key Concepts
Center of a CircleRadius of a CircleStandard Form
Center of a Circle
The center of a circle is a crucial concept when studying circle equations. In the standard form of a circle's equation, \((x-h)^2 + (y-k)^2 = r^2\), the center is represented by the coordinates \((h, k)\). To find the center:
For example, in the equation \((x+1)^2 + (y+6)^2 = 19\), setting \(x+1 = 0\) and \(y+6 = 0\) gives \(x = -1\) and \(y = -6\), so the center is \((-1, -6)\).
Finding the center is like finding the middle point of the circle where all points are equidistant.
- Identify the numbers inside the parentheses next to \(x\) and \(y\).
- Set those expressions equal to zero: \(x - h = 0\) and \(y - k = 0\).
For example, in the equation \((x+1)^2 + (y+6)^2 = 19\), setting \(x+1 = 0\) and \(y+6 = 0\) gives \(x = -1\) and \(y = -6\), so the center is \((-1, -6)\).
Finding the center is like finding the middle point of the circle where all points are equidistant.
Radius of a Circle
The radius of a circle is the distance from the center to any point on the circle. In the standard form equation:
For example, in the equation \((x+1)^2 + (y+6)^2 = 19\), the number \(19\) represents \(r^2\). Taking the square root, the radius \(r\) is \(\sqrt{19}\).
The radius is vital for understanding the size of the circle and is used in many applications, such as calculating circumference and area.
- The term \(r^2\) on the right side is the square of the radius.
- Simply take the square root of this number to find the radius \(r\).
For example, in the equation \((x+1)^2 + (y+6)^2 = 19\), the number \(19\) represents \(r^2\). Taking the square root, the radius \(r\) is \(\sqrt{19}\).
The radius is vital for understanding the size of the circle and is used in many applications, such as calculating circumference and area.
Standard Form
The standard form of a circle's equation simplifies the process of identifying its center and radius. This form is expressed as: \[(x-h)^2 + (y-k)^2 = r^2\]Here’s how to use it:
For instance, the equation \((x+1)^2 + (y+6)^2 = 19\) fits this format, allowing us to easily extract the center \((-1, -6)\) and the radius \(\sqrt{19}\).
Using the standard form makes these calculations straightforward, providing a clear framework for analyzing a circle's properties.
- \(h\) and \(k\) are the coordinates of the center.
- \(r\) is the radius, found by taking the square root of \(r^2\).
For instance, the equation \((x+1)^2 + (y+6)^2 = 19\) fits this format, allowing us to easily extract the center \((-1, -6)\) and the radius \(\sqrt{19}\).
Using the standard form makes these calculations straightforward, providing a clear framework for analyzing a circle's properties.
Other exercises in this chapter
Problem 17
(a) sketch the curve represented by the parametric equations (indicate the orientation of the curve). Use a graphing utility to confirm your result. (b) Elimina
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Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (2,3),(2,-3)\(;\) passes through the point (0,5)
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Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. Foci: (0,±5)\(;\) major axis of length 14
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