What is the standard form of the equation of a circle with endpoints of the diameter at (0,10) and (-10,-2)?

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Answer 1 · 2016-03-24T16:35:05

$(x + 5)^2 + (y - 4)^2 = 61$

The equation of a circle in standard form is

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

where
$h$ : $x$ -coordinate of the center
$k$ : $y$ -coordinate of the center
$r$ : radius of the circle

To get the center, get the midpoint of the endpoints of the diameter

#h = (x_1 + x_2)/2

$=> h = (0 + -10)/2$
$=> h = -5$

$k = (y_1 + y_2)/2$

$=> k = (10 + -2)/2$
$=> k = 4$

$c: (-5, 4)$

To get the radius, get the distance between the center and either endpoint of the diameter

$r = sqrt((x_1 - h)^2 + (y_1 - k)^2)$

$r = sqrt((0 - -5)^2 + (10 - 4)^2)$

$r = sqrt(5^2 + 6^2)$

$r = sqrt61$

Hence, the equation of the circle is

$(x - -5)^2 + (y - 4)^2 = (sqrt61)^2$

$=> (x + 5)^2 + (y - 4)^2 = 61$