Question

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

Answer 1

Answer:

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

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*