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Chapter 7: Problem 23

We can find an equation of a circle if we know the coordinates of the endpoints of a diameter of the circle. First, find the midpoint of the diameter, which is the center of the circle. Then find the radius, which is the distance from the center to either endpoint of the diameter. Finally use the center-radius form to find the equation. Find the center-radius form for each circle having the given endpoints of a diameter. $$(-5,-7) \text { and }(1,1)$$

Short Answer

Expert verified
The equation of the circle is \\( (x + 2)^2 + (y + 3)^2 = 25 \\).

Step by step solution

01

Find the Midpoint

To find the midpoint of a line segment given endpoints \((-5,-7)\) and \(1,1\), use the midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \). Substitute the given values: \[ \left( \frac{-5 + 1}{2}, \frac{-7 + 1}{2} \right) = \left( \frac{-4}{2}, \frac{-6}{2} \right) = (-2, -3) \]. Thus, the midpoint (and center of the circle) is \((-2,-3)\).
02

Calculate the Radius

To find the radius, first calculate the distance between the center \((-2, -3)\) and one of the endpoints, say \(1, 1\). Use the distance formula: \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. Substitute the values: \[ \sqrt{(1 - (-2))^2 + (1 - (-3))^2} = \sqrt{(1 + 2)^2 + (1 + 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]. The radius is 5.
03

Write the Equation in Center-Radius Form

The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius. We have the center \((-2, -3)\) and radius \5\. Substitute these into the equation:\((x + 2)^2 + (y + 3)^2 = 5^2\). Simplify the equation: \( (x + 2)^2 + (y + 3)^2 = 25 \). This is the equation of the circle.

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Key Concepts

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

Midpoint Formula
The midpoint formula is a handy tool for finding the middle point on a line segment between two points. If you have two endpoints of a diameter, like
  • Point A: \((-5, -7)\)
  • Point B: \((1, 1)\)
You use the formula: \[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]This involves adding the x-coordinates and y-coordinates separately, then dividing each by two:
  • First, \( \frac{-5 + 1}{2} = -2 \)
  • Then, \( \frac{-7 + 1}{2} = -3 \)
Thus, the midpoint is \((-2, -3)\). This point is also the center of the circle when dealing with a circle's diameter. Having the midpoint is crucial because it serves as our circle's center for writing the circle's equation.
Radius Calculation
Once the center of the circle is known, the next step is determining the radius, which is the distance from this center to any endpoint of the diameter. In our example, the center is at \((-2, -3)\), and we choose endpoint \((1, 1)\). To find this distance, we use the distance formula, which is:\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]For our particular points:
  • Calculate \((1 - (-2)) = 1 + 2 = 3\)
  • Calculate \((1 - (-3)) = 1 + 3 = 4\)
Thus, the distance is:\[\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]This 5-unit distance is the radius of our circle, showing how far the boundary of the circle extends from its center.
Distance Formula
The distance formula determines the straight-line distance between two points in a plane. It is vital in geometry, especially when calculating the size of geometric figures like circles. Given two points
  • \((x_1, y_1)\)
  • \((x_2, y_2)\)
the distance formula is:\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula works by creating a right triangle from the two points, with differences in x and y as the triangle's legs. The distance is then the hypotenuse. In our circle context, it confirms that the radius is 5 units. The distance formula is a general tool useful beyond circles, aiding any scenario requiring precise point-to-point distance measurement.
Center-Radius Form
The center-radius form is the standard way to express the equation of a circle. It emphasizes the circle's center coordinates and its radius, offering a straightforward description of the circle's size and location. The general form is:\[(x - h)^2 + (y - k)^2 = r^2\]
  • \((h, k)\) represents the circle's center coordinates
  • \(r\) stands for the circle's radius
For our specific circle:
  • Center \((-2, -3)\)
  • Radius \(5\)
The equation becomes:\[(x + 2)^2 + (y + 3)^2 = 25\]This equation represents all points that lie on the circle, each spaced exactly 5 units from the center. This form is easy to interpret and helps in visualizing or sketching the circle's position on a coordinate grid.

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Most popular questions from this chapter

Find an equation for each hyperbola. \(x\) -intercepts ( \(\pm 3,0\) ); foci ( \(\pm 4,0\) )

Find an equation of a parabola that satisfies the given conditions. Focus \((-1,2) ;\) vertex \((3,2)\)

Path of an Object on a Planet When an object moves under the influence of a gravitational force (without air resistance), its path can be parabolic. This is the path of a ball thrown near the surface of a planet or other celestial object. Suppose two balls are simultaneously thrown upward at a \(45^{\circ}\) angle on two different planets. If their initial velocities are both \(30 \mathrm{mph}\), then their \(x y\) -coordinates in feet can be expressed by the equation $$ y=x-\frac{g}{1922} x^{2} $$ where \(g\) is the acceleration due to gravity. The value of \(g\) will vary with the mass and size of the planet. (Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.) (a) On Earth, \(g=32.2\) and on Mars, \(g=12.6 .\) Find the two equations, and use the same screen of a graphing calculator to graph the paths of the two balls thrown on Earth and Mars. Use the window [0,180] by \([0,120] .\) (Hint: If possible, set the mode on your graphing calculator to simultaneous.) (b) Determine the difference in the horizontal distances traveled by the two balls.

Find an equation for each hyperbola. Vertices \((-3,0)\) and \((3,0) ;\) passing through \((6,1)\)

For individual or group investigation. Consider the ellipse and hyperbola defined by $$\frac{x^{2}}{16}+\frac{y^{2}}{12}=1 \quad \text { and } \quad \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$$ respectively. Graph the ellipse with your calculator, and trace to find the coordinates of several points on the ellipse.
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