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Chapter 8: Problem 18

Find the center-radius form for each circle satisfying the given conditions. Center \((5,-1) ;\) tangent to the \(y\) -axis

Short Answer

Expert verified
\((x - 5)^2 + (y + 1)^2 = 25\)

Step by step solution

01

Identify the Circle's Center and Tangency

The center of the circle is given as \((5, -1)\). Since the circle is tangent to the \(y\)-axis, the distance from the center to the \(y\)-axis equals the radius.
02

Determine the Radius

Since the circle is tangent to the \(y\)-axis at 5 units away from \(x = 0\), the radius \(r\) must be 5 units.
03

Write the Equation in Center-Radius Form

The center-radius form of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center. Here, the center is \((5, -1)\) and the radius \(r = 5\). Substitute these into the equation to get:\[(x - 5)^2 + (y + 1)^2 = 25\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Center-Radius Form
The center-radius form of a circle is a very useful way to write the equation of a circle. It gives clear information about the circle's geometry. The equation looks like this:
  • \((x - h)^2 + (y - k)^2 = r^2\)
Here,
  • \(h\) and \(k\) are the coordinates of the circle's center,
  • \(r\) is the radius of the circle.
This form explicitly displays the circle's center and its radius squared. It makes visualizing and graphing circles more straightforward and helps in solving geometry-related problems with precision. For instance, if you know the center of a circle is at \((5, -1)\) and you determine the radius to be 5, you can easily write its equation using this form.
Tangent to Axis
When a circle is tangent to an axis, it means it just touches the axis at a single point. This tangency provides useful geometric information. In our scenario, the circle is tangent to the \(y\)-axis. Since it is tangent to the \(y\)-axis, the horizontal distance from its center point to the \(y\)-axis is equal to the circle's radius. Furthermore, this implies that the circle neither crosses nor penetrates the axis at any other point.Here are a few key points to remember:
  • The point of tangency occurs where the circle just touches but does not intersect the axis.
  • This tangency point gives us a direct measurement of the radius, seen as the distance from the center to the axis.
  • Tangency simplifies calculations since it involves straightforward distance measurements between the center and the axis.
Radius Determination
Determining the radius of a circle that is tangent to an axis can be relatively straightforward. In situations where the circle is tangent to, for example, the \(y\)-axis, the radius is simply the horizontal distance from the circle鈥檚 center to the \(y\)-axis itself.For our given example, the center of the circle is at \((5, -1)\). The \(x\)-coordinate
  • 5 units away from the \(y\)-axis which means the radius is 5 units.
This is because the circle's edge just touches the axis at that distance, providing a clear and accurate measurement for the circle's radius:
  • The radius is a key component in constructing the circle's equation since it helps define the shape and size of the circle in relation to its center.
  • Knowing the radius supports further mathematical operations and graphing tasks linked to the circle.
Equation Formulation
To formulate the equation of a circle, especially when given a center and a condition like being tangent to an axis, involves substituting known values into the center-radius form. Given the circle's center at \((5, -1)\), and radius \(5\), the formulation proceeds as follows:Start with the standard center-radius form:
  • \((x-h)^2 + (y-k)^2 = r^2\)
Substitute the center coordinates \(h = 5\) and \(k = -1\), and the radius \(r = 5\):
  • \((x-5)^2 + (y+1)^2 = 25\)
Here's how to validate each step:
  • Ensure correct substitution of center coordinates: \((h, k) = (5, -1)\).
  • Calculate the radius square correctly: \(r^2 = 25\).
  • This yields the final equation of the circle that fits the given conditions.
This formulation helps consolidate the circle's location and size graphically and algebraically, making it essential for many geometry tasks.

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