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Write the standard equation of a circle having endpoints of a diameter at (-1, 6) and (9,-8).

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[tex]r ^{2} = (x-4)^2 + (y+1)^{2}[/tex]

Given:

endpoints at (-1, 6) and (9, -8)

Solve for:

the standard equation of a circle having endpoints of a diameter at (-1, 6) and (9,-8).

Step-by-step explanation:

[tex]d=\sqrt{(x_{2} -x_{1})^2+(y_{2} -y_{1})^2}[/tex]

[tex]d=\sqrt{(9-(-1))^2+(-8-6)^2}[/tex]

[tex]d=\sqrt{10^2+-14^2}[/tex]

[tex]d=\sqrt{100+196}[/tex]

[tex]d=\sqrt{296}[/tex]

[tex]d=2\sqrt{74}[/tex]

[tex]r=\frac{d}{2}[/tex]

[tex]r=\frac{2\sqrt{74} }{2}[/tex]

[tex]r=\sqrt{74}[/tex]

[tex]Center = (\frac{x_{1}+x_{2} }{2} ,\frac{y_{1}+y_{2} }{2})[/tex]

[tex]= (\frac{-1+9}{2}, \frac{6+(-8)}{2} )[/tex]

[tex]=(4, -1)[/tex]

[tex]Center = (4,-1) \\and\\r = \sqrt{74}[/tex]

Equation:

[tex](\sqrt{74}) ^{2} = (x-4)^2 + (y+1)^{2}[/tex]

[tex]r ^{2} = (x-4)^2 + (y+1)^{2}[/tex]

*This took forever so I hoped it helped lol*
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