Find point of Equidistance

[arizona_cards_11]

Find point of Equidistance...

Find the point that is equidistant from (0,0) , (2,3) , and (3,-2)...

I have plotted the points and have drawn a circle, putting all points on the outside...showing the equidistance.

I plan on guesstimating a point int he middle, labeling it (X1,Y1)...then, I will plug each of the points above into the distance formula seperately with the "X1/Y1" creating 3 equations.

Then, I'll use substitution and such to find x and y...does this seem like a logical plan?

 

[Robokapp]

Your plan is what I first looked at but then I figured why not try distance formula, maybe i can learn something about the triangle this forms.

(0, 0) => (3, -2) = $$\sqrt{13}$$ units.
(0, 0) => (2, 3) = $$\sqrt{13}$$ units.
(2, 3) => (3, -2) = $$\sqrt{26}$$ units = $$\sqrt{13}*\sqrt{2}$$

Without looking at the graph i can tell your points form an isosceles right triangle with vertice in (0, 0). Maybe that's useful. Now inscribe that in a circle...and remember some basic geometry properties.

Edit: It doesn't look like Calculus to me...is there a fancy way to do this?

 

[arizona_cards_11]

Your plan is what I first looked at but then I figured why not try distance formula, maybe i can learn something about the triangle this forms.

(0, 0) => (3, -2) = $$\sqrt{13}$$ units.
(0, 0) => (2, 3) = $$\sqrt{13}$$ units.
(2, 3) => (3, -2) = $$\sqrt{26}$$ units = $$\sqrt{13}*\sqrt{2}$$

Without looking at the graph i can tell your points form an isosceles right triangle with vertice in (0, 0). Maybe that's useful. Now inscribe that in a circle...and remember some basic geometry properties.

Edit: It doesn't look like Calculus to me...is there a fancy way to do this?

It is trig in a Calculus book...I should have put it into a different forum, sorry.

 

[Robokapp]

No lol I was just wondering...because I used nothing but Distance Formula and basic Geometry I info...and I got curious if I either did a big mistake or I found a "lucky" numerical situation.

Like...any 4th grader can do 2^2 but when you ask 2^3 there will be some answers of 5 or 6. It happens to me a lot to "mistake" into the right answer.

 

[arizona_cards_11]

The site that has helped me the most is the following link...

http://www.regentsprep.org/regents/math/geometry/GCG6/RCir.htm

 

[HallsofIvy]

Given any triangle, its "circumcenter" is the center of the circumscribing circle- the circle that goes through the three vertices. Obviously, the circumcenter is equidistant from the three vertices.

Since the circumcenter is equidistant from the two endpoints of any side, it must lie on the perpendicular bisector of that side. In other words, the circumcenter is at the intersection of the perpendicular bisectors of the three sides.

In the given example, vertices at (0,0) , (2,3) , and (3,-2):
The side between (0,0) and (2,3) has slope 3/2 and midpoint (1, 3/2). It's perpendicular bisector has slope -2/3 and passes through (1, 3/2) and so has equation y= (-2/3)(x- 1)+ 3/2.
The side between (0,0) and (3,-2) has slope -2/3 and midpoint (3/2, -1). It's perpendicular bisector has slope 3/2 and passes through (3/2, -1) and so has equation y= (3/2)(x- 3/2)- 1.

The circumcenter is the point of intersection of those two lines: Solve the two equations for x and y.

As a check, the perpendicular bisector of the third side through (2, 3) and (3, -2) should also pass through that point.