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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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$$r ^{2} = (x-4)^2 + (y+1)^{2}$$

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:

$$d=\sqrt{(x_{2} -x_{1})^2+(y_{2} -y_{1})^2}$$

$$d=\sqrt{(9-(-1))^2+(-8-6)^2}$$

$$d=\sqrt{10^2+-14^2}$$

$$d=\sqrt{100+196}$$

$$d=\sqrt{296}$$

$$d=2\sqrt{74}$$

$$r=\frac{d}{2}$$

$$r=\frac{2\sqrt{74} }{2}$$

$$r=\sqrt{74}$$

$$Center = (\frac{x_{1}+x_{2} }{2} ,\frac{y_{1}+y_{2} }{2})$$

$$= (\frac{-1+9}{2}, \frac{6+(-8)}{2} )$$

$$=(4, -1)$$

$$Center = (4,-1) \\and\\r = \sqrt{74}$$

Equation:

$$(\sqrt{74}) ^{2} = (x-4)^2 + (y+1)^{2}$$

$$r ^{2} = (x-4)^2 + (y+1)^{2}$$

*This took forever so I hoped it helped lol*
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