How do you write the standard form of the equation of the circle whose diameter has endpoints of (-2, 4) and (4, 12)?

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Answer 1 · 2016-07-31T06:51:12

$(x-1)^2+(y-8)^2=25$

The given data are the endpoints $E_1(x_1, y_1)=(-2, 4)$ and $E_2(x_2, y_2)=(4, 12)$ of the diameter $D$ of the circle

Solve for the center $(h, k)$

$h=(x_1+x_2)/2=(-2+4)/2=1$
$k=(y_1+y_2)/2=(4+12)/2=8$

Center $(h, k)=(1, 8)$

Solve now for the radius $r$

$r=D/2=(sqrt((x_1-x_2)^2+(y_1-y_2)^2))/2$

$r=D/2=(sqrt((-2-4)^2+(4-12)^2))/2$

$r=D/2=sqrt(36+64)/2$

$r=D/2=sqrt(100)/2$

$r=D/2=10/2$

$r=5$

The standard form of the equation of the circle:

Center-Radius Form

$(x-h)^2+(y-k)^2=r^2$

$(x-1)^2+(y-8)^2=5^2$

$(x-1)^2+(y-8)^2=25$

God bless....I hope the explanation is useful.