Answer: [tex](x - 6)^2 + (y + 1)^2 = 40[/tex]
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Explanation:
The x coordinates of the given endpoints are x = 8 and x = 4. Apply the midpoint to them to get [tex]\frac{x_1+x_2}{2} = \frac{8+4}{2} = \frac{12}{2} = 6[/tex] which is the x coordinate of the midpoint. I added them up and divided by 2.
Similar steps will have the y coordinate of the midpoint be
[tex]\frac{y_1+y_2}{2} = \frac{-7+5}{2} = \frac{-2}{2} = -1[/tex]
Overall, the midpoint of those given points is (6, -1)
This is the center of the circle because the two endpoints are part of a diameter.
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Now turn to the template
[tex](x - h)^2 + (y -k )^2 = r^2[/tex]
We will plug in the center (h,k) = (6, -1) and one of the endpoints for (x,y). Let's say we go for (x,y) = (4,5)
This leads us to the following:
[tex](x - h)^2 + (y -k )^2 = r^2\\\\(4 - 6)^2 + (5 - (-1) )^2 = r^2\\\\(4 - 6)^2 + (5 + 1 )^2 = r^2\\\\(-2)^2 + (6 )^2 = r^2\\\\4 + 36 = r^2\\\\r^2 = 40\\\\[/tex]
We don't need to isolate r itself. If you wanted, you could take the square root of both sides to see that the radius is [tex]r = \sqrt{40} \approx 6.3246[/tex]. However, again we don't need to solve for r entirely when forming the circle's equation.
With that [tex]r^2[/tex] value in mind, and the h & k values as well, we go from this template
[tex](x - h)^2 + (y -k )^2 = r^2[/tex]
to this final answer
[tex](x - 6)^2 + (y + 1)^2 = 40[/tex]
