use the following information to create an equation for a circle in standard form
PLEASE DO THIS WITHOUT A CALCULATOR. answer must match one of the answer choices. there is only one right answer. if your answer is none of these please check your math that answer choice is wrong.
Transcribed Image Text:**58. Endpoints of diameter:** \((3, 0)\) and \((7, 10)\)
In this problem, the endpoints of the diameter of a circle are given as the coordinates \((3, 0)\) and \((7, 10)\). To find the center of the circle, which is the midpoint of the diameter, you can calculate the average of the x-coordinates and the y-coordinates of the endpoints.
The midpoint formula is:
\[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the given values:
\[ \left(\frac{3 + 7}{2}, \frac{0 + 10}{2}\right) = \left(\frac{10}{2}, \frac{10}{2}\right) = (5, 5) \]
Thus, the center of the circle is \((5, 5)\).
Additionally, to calculate the radius, use the distance formula to find the length of the diameter, then divide by two:
The distance formula is:
\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the given values:
\[ \sqrt{(7 - 3)^2 + (10 - 0)^2} = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} \]
The length of the diameter is \(\sqrt{116}\), so the radius is:
\[ \frac{\sqrt{116}}{2} = \frac{2\sqrt{29}}{2} = \sqrt{29} \]
This explanation can be accompanied by a graph showing the circle, the diameter with endpoints \((3, 0)\) and \((7, 10)\), and the center at \((5, 5)\).
Transcribed Image Text:**Problem 8.2.58**
Consider the following equations and determine which one represents a circle centered at the point (5, 5):
A. \((x - 5)^2 + (y - 5)^2 = 1\)
B. \((x - 5)^2 + (y - 5)^2 = 116\)
C. \((x - 5)^2 + (y - 5)^2 = \sqrt{116}\)
D. \((x + 2)^2 + (y + 5)^2 = 29\)
E. \((x - 5)^2 + (y - 5)^2 = 29\)
Two-dimensional figure measured in terms of radius. It is formed by a set of points that are at a constant or fixed distance from a fixed point in the center of the plane. The parts of the circle are circumference, radius, diameter, chord, tangent, secant, arc of a circle, and segment in a circle.
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![**58. Endpoints of diameter:** \((3, 0)\) and \((7, 10)\)
In this problem, the endpoints of the diameter of a circle are given as the coordinates \((3, 0)\) and \((7, 10)\). To find the center of the circle, which is the midpoint of the diameter, you can calculate the average of the x-coordinates and the y-coordinates of the endpoints.
The midpoint formula is:
\[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the given values:
\[ \left(\frac{3 + 7}{2}, \frac{0 + 10}{2}\right) = \left(\frac{10}{2}, \frac{10}{2}\right) = (5, 5) \]
Thus, the center of the circle is \((5, 5)\).
Additionally, to calculate the radius, use the distance formula to find the length of the diameter, then divide by two:
The distance formula is:
\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the given values:
\[ \sqrt{(7 - 3)^2 + (10 - 0)^2} = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} \]
The length of the diameter is \(\sqrt{116}\), so the radius is:
\[ \frac{\sqrt{116}}{2} = \frac{2\sqrt{29}}{2} = \sqrt{29} \]
This explanation can be accompanied by a graph showing the circle, the diameter with endpoints \((3, 0)\) and \((7, 10)\), and the center at \((5, 5)\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdaa77415-d220-4776-9d7c-d055c20faba9%2F89ad3b49-411f-415a-ae0e-1c81146a5292%2Fprlq5aci_processed.png&w=3840&q=75)
