How to find the center of a circle given the three points (-3,5); (3,3); (11,19)?

May 17, 2016

The perpendicular bisectors of two secant lines intersect at the center.

Explanation:

Find the slopes of the lines through two pairs of points.
Find the midpoints of those lines

The perpendicular bisectors are the lines through the midpoints with slopes equal to the negative reciprocals of the slopes of the secant lines.
Find the equations of the perpendicular bisectors.

Now find the intersection of the lines whose equations you just found.

The secant joining $\left(- 3 , 5\right)$ and $\left(3 , 3\right)$ has slope $- \frac{1}{3}$ and midpoint $\left(0 , 4\right)$.
The equation of the perpendicular bisector of the secant is $y = 3 x + 4$

It is left to the reader to verify that the secnt jining $\left(3 , 3\right)$ and $\left(11 , 19\right)$ has perpendicular bisector $2 y = - x + 29$

The lines

$y = 3 x + 4$ and
$2 y = - x + 29$

intersect at $\left(3 , 13\right)$.

May 17, 2016

$\left(3 , 13\right)$
The center (a, b) is at the same distance = radius, from each of the three points on the circle. This gives a = 3 and b = 13..

Explanation:

The points are equidistant from the center (a, b). Equating squares of the distances,

${\left(a + 3\right)}^{2} + {\left(b - 5\right)}^{2} = {\left(a - 3\right)}^{2} + {\left(b - 3\right)}^{2} = {\left(a - 11\right)}^{2} + {\left(b - 19\right)}^{2}$,

From the first two,

$3 a - b = 4$, and from first and the third,

$a + b = 16$.

solving, (a, b) = (3, 13)...