Answer:
[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*
