MHB Finding the Equation of a Circle Passing Through Three Given Points

[karush]

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.

 

[skeeter]

find an equation of the circle passing through the given points
$(-1,3),\quad (6,2),\quad (-2,-4)$

slope between (-1,3) and (-2,-4) ... $m = 7$

midpoint between (-1,3) and (-2,-4) ... (-3/2, -1/2)

perpendicular bisector ... $y + \dfrac{1}{2} = -\dfrac{1}{7}\left(x + \dfrac{3}{2}\right)$
slope between (-1,3) and (6,2) ... $m = -\dfrac{1}{7}$

midpoint between (-1,3) and (6,2) ... (5/2, 5/2)

perpendicular bisector ... $y - \dfrac{5}{2} = 7 \left(x - \dfrac{5}{2} \right)$intersection of the two perpendicular bisectors is the circle center, (2, -1)distance between any point and the center is the radius ... $r = 5$

circle equation ...

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

 

[karush]

well that was certainly very novel..

most of these examples throw you to the general equations and do a lot of gymnastics.

ok I assume that is a Desmos graph...

 

[topsquark]

slope between (-1,3) and (-2,-4) ... $m = 7$

midpoint between (-1,3) and (-2,-4) ... (-3/2, -1/2)

perpendicular bisector ... $y + \dfrac{1}{2} = -\dfrac{1}{7}\left(x + \dfrac{3}{2}\right)$
slope between (-1,3) and (6,2) ... $m = -\dfrac{1}{7}$

midpoint between (-1,3) and (6,2) ... (5/2, 5/2)

perpendicular bisector ... $y - \dfrac{5}{2} = 7 \left(x - \dfrac{5}{2} \right)$intersection of the two perpendicular bisectors is the circle center, (2, -1)distance between any point and the center is the radius ... $r = 5$

circle equation ...

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

Well, well, well! Someone's been reading their Euclid! (Bow)

Nice job.

-Dan

 

[skeeter]

well that was certainly very novel..

most of these examples throw you to the general equations and do a lot of gymnastics.

ok I assume that is a Desmos graph...

not desmos ... free graphing program for windows (maybe there is one for Macs, haven't looked to confirm)https://www.padowan.dk/download/

 

[karush]

I think I'm mostly interested in what ca be done with tikx.
Albeit with its limitations

 

[Greg]

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

$$(-1-h)^2-(6-h)^2+(3-k)^2-(2-k)^2=0$$

$$-7(5-2h)+(5-2k)=0$$

$$-35+14h+5-2k=0$$

$$-30+14h-2k=0\quad[1]$$

$$(-1-h)^2-(-2-h)^2+(3-k)^2-(-4-k)^2=0$$

$$(-3-2h)+7(-1-2k)=0$$

$$-10-2h-14k=0\quad[2]$$

$$[1]+7*[2]\implies k=-1\implies h=2$$

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

 

[karush]

so I quess there is no slam dunk one step solution :(