Answer to Question #91105 in Analytic Geometry for King

Question #91105
An archeologist found the remains of an ancient wheel, which she then placed on a grid. If an arc of the wheel passes through A(–7, 0), B(–3, 4) and C(7, 0), locate the center of the wheel, and the standard equation of the circle defining its boundary.
1
Expert's answer
2019-06-24T11:15:56-0400

The standard equation of the circle defining its boundary is


"(x-a)^2 + (y-b)^2 = R^2"

where the point with the coordinates (a, b) is the center of the circle. Put each point in the equation. Then we get the system of equations:


"\\left\\{\n\\begin{array}{rcl}\n{(-7-a)^2 + (0 - b)^2 = R^2,} \\\\\n{(-3-a)^2 + (4 - b)^2 = R^2,} \\\\\n{(7-a)^2 + (0 - b)^2 = R^2.}\n\\end{array}\n\\right."

Open brackets (use the formula "(a+b)^2 = a^2 + b^2 + 2ab") :


"\\left\\{\n\\begin{array}{rcl}\n{49 +a^2+ 14a + b^2 = R^2,} \\\\\n{9+a^2 +6a+ 16+ b^2 - 8b = R^2,} \\\\\n{49 +a^2- 14a + b^2 = R^2.}\n\\end{array}\n\\right."

Subtract from the first equation the third:

"49 +a^2+ 14a + b^2 - (49 +a^2- 14a + b^2) = R^2 -R^2"

"28a = 0 \\implies a = 0"

Equate the first equation to the second:

"49 +a^2+ 14a + b^2 = 9+a^2 +6a+ 16+ b^2 - 8b"

Because "a = 0":

"49 +b^2 = 9+16+ b^2 - 8b"

"25 - 8b = 49\n-8b = 24 \\implies b = -3"

Then the center of the circle is (0, -3).

Find the radius of the circle: put point and the coordinates of center in the equation:

"(-7-0)^2 + (0 - (-3))^2 = R^2"

"49 +9 = R^2 \\implies R =\\sqrt{58}"

The standard equation of the circle:


"x^2 + (y+3)^2 = 58"


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