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Chapter 1: Problem 43

In Exercises \(41-46,\) find an equation for the circle with the given center \(C(h, k)\) and radius \(a\) . Then sketch the circle in the \(x y\) -plane. Include the circle's center in your sketch. Also, label the circle's \(x\) - and \(y\) -intercepts, if any, with their coordinate pairs. $$ C(-1,5), \quad a=\sqrt{10} $$

Short Answer

Expert verified
The circle's equation is \((x + 1)^2 + (y - 5)^2 = 10\). Intercepts: x at \((-6, 0), (4, 0)\); y at \((0, 8), (0, 2)\).

Step by step solution

01

Understanding the Standard Form of Circle Equation

The standard equation for a circle with center \( C(h, k) \) and radius \( a \) is \( (x - h)^2 + (y - k)^2 = a^2 \). This formula describes every point \((x, y)\) that is at a distance \( a \) from the center \( C(h, k) \).
02

Plug in the Center Coordinates and Radius

Given the center \( C(-1, 5) \) and radius \( a = \sqrt{10} \), substitute \( h = -1 \), \( k = 5 \), and \( a = \sqrt{10} \) into the standard circle equation. The equation becomes: \((x - (-1))^2 + (y - 5)^2 = (\sqrt{10})^2\).
03

Simplify the Circle Equation

Solve the equation from Step 2. First, \(x - (-1)\) simplifies to \(x + 1\), and \((\sqrt{10})^2\) becomes 10. Thus, the equation is \((x + 1)^2 + (y - 5)^2 = 10\).
04

Sketch the Circle

Draw the circle in the \(xy\)-plane: Center at \((-1, 5)\) and a radius of \(\sqrt{10}\), about 3.16 units. Include the circle's center and note the intercepts. Calculate x- and y-intercepts by setting the respective variables to zero in the equation \((x+1)^2 + (y-5)^2 = 10\).
05

Find the X-intercepts

Set \(y = 0\) in the equation: \((x+1)^2 + (0-5)^2 = 10\) which simplifies to \((x+1)^2 = 25\). Solving for \(x\), we get \(x = -6\) or \(x = 4\). So, the x-intercepts are \((-6, 0)\) and \((4, 0)\).
06

Find the Y-intercept

Set \(x = 0\) in the equation: \((0+1)^2 + (y-5)^2 = 10\) which simplifies to \(1 + (y-5)^2 = 10\) and \((y-5)^2 = 9\). Solving for \(y\), we have \(y = 8\) or \(y = 2\). Thus, y-intercepts are \((0, 8)\) and \((0, 2)\).
07

Label intercepts on the Sketch

On your sketch, clearly mark the x-intercepts \((-6, 0)\) and \((4, 0)\), and the y-intercepts \((0, 8)\) and \((0, 2)\). Clearly label the center of the circle at \((-1, 5)\).

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Key Concepts

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

Standard Form
When studying circles in geometry, the standard form of its equation is a key tool. The standard form of a circle's equation is written as \[(x - h)^2 + (y - k)^2 = a^2\] where
  • \(h\) and \(k\) represent the \(x\) and \(y\) coordinates of the circle's center, respectively.
  • \(a\) is the radius of the circle.
This equation represents all the points \((x, y)\) that comprise the circle. Essentially, it expresses that every point on the circle is exactly a distance \(a\) from its center \((h, k)\). With this form, you can see easily both where the center of the circle is located and how large the circle is. In practice, you replace \(h\), \(k\), and \(a\) with specific values to derive the actual equation of a circle.
Center of a Circle
The center of a circle is crucial because it is the point from which all points on the circle are equidistant. In the equation \((x - h)^2 + (y - k)^2 = a^2\), the center is represented by the coordinates \((h, k)\). For instance, if the center is \((-1, 5)\), inserting these values into the equation shows precisely where the circle is situated on the coordinate plane. Thus, the modified equation becomes \((x + 1)^2 + (y - 5)^2 = a^2\). Recognizing where the center is helps you not only in drawing the circle but also in quickly understanding its orientation relative to the axes.
Radius of a Circle
The radius of a circle is the distance from the center to any point on the circle's edge. It is denoted by \(a\) in the standard equation \((x - h)^2 + (y - k)^2 = a^2\). For example, if \(a = \sqrt{10}\), the circle's radius measures about 3.16 units, as the square root of 10 is approximately 3.16. The larger the radius, the larger the circle will be. In this equation example, substituting \(a = \sqrt{10}\) simplifies to \(10\) when squared, giving the complete circle equation as:\[(x + 1)^2 + (y - 5)^2 = 10\]Understanding the radius allows you to construct the circle and recognize how tightly or loosely it fits around its defined center.
X-Intercepts
To find where a circle crosses the x-axis, you need to calculate the x-intercepts. These occur when the \(y\)-value is zero. Plugging \(y = 0\) into the equation \((x + 1)^2 + (y - 5)^2 = 10\) gives:\((x + 1)^2 + 25 = 10\).Solving the equation leads to:\((x+1)^2 = 25\),which simplifies to two possible solutions for \(x\):
  • \(x = -6\)
  • \(x = 4\)
Thus, the x-intercepts of this circle are at points \((-6, 0)\) and \((4, 0)\). These intersections tell you exactly where the circle meets the x-axis.
Y-Intercepts
Y-intercepts are where the circle intersects the y-axis. To find these, set \(x = 0\) in the equation \((x + 1)^2 + (y - 5)^2 = 10\). This results in:\(1 + (y - 5)^2 = 10\),which simplifies to \((y - 5)^2 = 9\).Solving for \(y\), we find:
  • \(y = 8\)
  • \(y = 2\)
Thus, the y-intercepts are \((0, 8)\) and \((0, 2)\). Knowing these y-intercepts allows you to understand where the circle passes through the y-axis, which is critical in visualizing its placement and crossing points on the plane.

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Most popular questions from this chapter

In Exercises \(67-70,\) you will explore graphically the general sine function $$f(x)=37 \sin \left(\frac{2 \pi}{365}(x-101)\right)+25$$ as you change the values of the constants \(A, B, C,\) and \(D .\) Use a CAS or computer grapher to perform the steps in the exercises. The vertical shift \(D \quad\) Set the constants \(A=3, B=6, C=0\) a. Plot \(f(x)\) for the values \(D=0,1,\) and 3 over the interval \(-4 \pi \leq x \leq 4 \pi\) . Describe what happens to the graph of the general sine function as \(D\) increases through positive values. b. What happens to the graph for negative values of \(D ?\)

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