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Answer:
[tex]x^2 + y^2 - 40 = 0[/tex]
Step-by-step explanation:
Given
[tex](h_1,k_1) = (6,-2)[/tex]
[tex](h_2,k_2) = (-6,2)[/tex]
Required
Determine the equation of the circle
First, calculate the midpoint of the given coordinates:
[tex](h,k) = \frac{1}{2}(h_1+h_2,k_1+k_2)[/tex]
[tex](h,k) = \frac{1}{2}(6-6,-2+2)[/tex]
[tex](h,k) = \frac{1}{2}(0,0)[/tex]
[tex](h,k) = (0,0)[/tex]
Next, calculate the radius.
This is the distance between [tex](0,0)[/tex] calculated above and any of [tex](6,-2)[/tex] [tex](-6,2)[/tex]
Using:
[tex](0,0)[/tex] and [tex](-6,2)[/tex], the radius is:
[tex]r = \sqrt{(0 - (-6))^2 + (0-2)^2}[/tex]
[tex]r = \sqrt{40}[/tex]
The equation is then calculated using:
[tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Where:
[tex]r = \sqrt{40}[/tex] and [tex](h,k) = (0,0)[/tex]
[tex](x - 0)^2 + (y - 0)^2 = (\sqrt{40})^2[/tex]
[tex](x )^2 + (y)^2 = 40[/tex]
[tex]x^2 + y^2 = 40[/tex]
[tex]x^2 + y^2 - 40 = 0[/tex]