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#### karush

##### Well-known member

- Jan 31, 2012

- 2,685

find an equation of the circle passing through the given points

85 Given $(-1,3),\quad (6,2),\quad (-2,-4)$

since the radius is the same for all points set all cirlce eq equal to each other

$(x_1-h)^2+(y_1-k)^2=(x_2-h)^2+(y_2-k)^2=(x_3-h)^2+(y_3-k)^2$

plug in values

$(-1-h)^2+(3-k)^2=(6-h)^2+(2-k)^2=(-2-h)^2+(-4-k)^2$

from this we get (via W|A)

$h = 2,\quad k = -1$

derive the radius by the distance of the center to one of the points

$d=\sqrt{(2-(-2))^2+(-1-(-4))^2}\sqrt{16+9}=\sqrt{25}=5$

thus the stardard circle equation would be

$\left(x-2\right)^2+\left(y+1\right)^2=25$

ok I think this is ok, but I was going to try to do this with a matrix but with the squares in it didn't see how

also want to try to draw the 3 points, center, and circle with tikx

first attempt... need 3 points and ticks and text

\begin{tikzpicture}

\draw (.4,-.2) circle (1cm);

\draw (-2,0) -- (3,0);

\draw (0,-2) -- (0,2);

\end{tikzpicture}

I assume there might be some kill all formula used for this problem.

85 Given $(-1,3),\quad (6,2),\quad (-2,-4)$

since the radius is the same for all points set all cirlce eq equal to each other

$(x_1-h)^2+(y_1-k)^2=(x_2-h)^2+(y_2-k)^2=(x_3-h)^2+(y_3-k)^2$

plug in values

$(-1-h)^2+(3-k)^2=(6-h)^2+(2-k)^2=(-2-h)^2+(-4-k)^2$

from this we get (via W|A)

$h = 2,\quad k = -1$

derive the radius by the distance of the center to one of the points

$d=\sqrt{(2-(-2))^2+(-1-(-4))^2}\sqrt{16+9}=\sqrt{25}=5$

thus the stardard circle equation would be

$\left(x-2\right)^2+\left(y+1\right)^2=25$

ok I think this is ok, but I was going to try to do this with a matrix but with the squares in it didn't see how

also want to try to draw the 3 points, center, and circle with tikx

first attempt... need 3 points and ticks and text

\begin{tikzpicture}

\draw (.4,-.2) circle (1cm);

\draw (-2,0) -- (3,0);

\draw (0,-2) -- (0,2);

\end{tikzpicture}

I assume there might be some kill all formula used for this problem.

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