Finding the Equation of a Circle Passing Through Three Given Points

  • Context: MHB 
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Circle Points
Click For Summary

Discussion Overview

The discussion centers on finding the equation of a circle that passes through three specified points: $(-1,3)$, $(6,2)$, and $(-2,-4)$. Participants explore various methods to derive the equation, including algebraic approaches and geometric reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the equality of distances from the center to the points, leading to the equation $(x_1-h)^2+(y_1-k)^2=(x_2-h)^2+(y_2-k)^2=(x_3-h)^2+(y_3-k)^2$ and derives the center $(h, k) = (2, -1)$ and radius $r = 5$.
  • Another participant proposes calculating the slopes and midpoints of the segments formed by the points to find the perpendicular bisectors, ultimately arriving at the same center and radius, but presents the circle equation as $(x-2)^2 + (y+1)^2 = 5^2$.
  • There is a mention of a desire to explore a matrix approach to the problem, indicating a search for alternative methods.
  • Some participants express surprise at the novelty of the methods discussed, contrasting them with more traditional approaches that rely heavily on general equations.
  • One participant expresses interest in using TikZ for graphical representation, acknowledging its limitations.
  • Another participant reflects on the complexity of the problem, suggesting that there is no straightforward solution.

Areas of Agreement / Disagreement

Participants generally agree on the center and radius of the circle, but there is no consensus on the best method to derive the equation or whether a simpler solution exists. Multiple approaches are presented, indicating a variety of perspectives on the problem.

Contextual Notes

Some mathematical steps and assumptions are not fully resolved, particularly regarding the use of matrices and the potential for a one-step solution. The discussion reflects a range of methods and interpretations without reaching a definitive conclusion on the most effective approach.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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.
 
Last edited:
Mathematics news on Phys.org
karush said:
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 ...

$(x-2)^2 + (y+1)^2 = 5^2$
 
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...
 
skeeter said:
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
 
karush said:
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/
 
I think I'm mostly interested in what ca be done with tikx.
Albeit with its limitations
 
$$(-1-h)^2+(3-k)^2=(6-h)^2+(2-k)^2=(-2-h)^2+(-4-k)^2$$

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

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

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

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

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

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

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

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

$$(x-2)^2+(y+1)^2=25$$
 
so I quess there is no slam dunk one step solution :(
 

Similar threads

MHB What are the coordinates of points equidistant from (1,0) and (5,4)?
  • · Replies 6 ·
Replies
6
Views
1K
MHB Finding the Equation of a Line through a Point and Circle Center?
  • · Replies 2 ·
Replies
2
Views
1K
MHB How Do You Find the Equation of a Line Through a Point and a Circle's Center?
  • · Replies 2 ·
Replies
2
Views
2K
MHB Equation of Circle in Standard Form
  • · Replies 6 ·
Replies
6
Views
2K
MHB How to Determine the Equation of a Circle from Given Diameter Coordinates?
  • · Replies 8 ·
Replies
8
Views
2K
MHB [ASK] A Line Intercepting A Circle
  • · Replies 4 ·
Replies
4
Views
1K
MHB Equation and Point Check for Circle of Radius 1
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Equation for circle points in 3D
  • · Replies 3 ·
Replies
3
Views
3K
MHB What is the equation of a circle tangent to the x-axis with a center at (3, 5)?
  • · Replies 2 ·
Replies
2
Views
1K
MHB How Do You Determine If a Point Lies on a Circle?
  • · Replies 4 ·
Replies
4
Views
2K