Parametric equation of a circle intersecting 3 points

[whitejac]

Homework Statement

Given the points P0 = (0,a), P1 = (b,0), P2 = (0,0), write the parametric equation of a circle that intersects the 3 points.

Assume that b > a and both are positive.

Homework Equations

X = h + rcos(t)
Y = k + rsin (t)
r = √((x-h)² + (y-k)²
Cos (t) = (x-h)/r
Sin (t) = (y-k) / r

The Attempt at a Solution

I'm having a really hard time visualizing parametric equations. I drew the circle and know that the challenge is finding the center, but I don't know how to express it. the circle is shifted right and up. I don't know how to express the coordinates of the center in terms of P0 and P1 as well as so it would intersect both these two points as well as (0,0).

 

[S.G. Janssens]

The three given points $P_0$, $P_1$ and $P_2$ are known. They all have the same unknown distance $r > 0$ to the unknown center $(h,k)$. Can you use this to set up and solve a system of equations in these unknowns?

 

[BvU]

A useful property of a circle is that the center point is the same distance from all points on the rim. So where are the points that have the same distance from (0,0) and (b,0) ? same for (0,0) and (0,a) ?

 

[whitejac]

Would it look something like this..?

At the origin:
R = √(h² + k²)

At point (b,0):
R =√(b-h)² + k²)

At point (0,a):
R = √(a-k)² + h²)

 

[BvU]

Yes. Not just 'something like' : that's what it is, exactly (*)
Now solve.

(*) assuming case insensitivity -- or did you have a purpose changing from r to R ?

 

[whitejac]

So...

k = √(r² - h²)

implies

r² = (b-h)² + r² - h²

Which reduces to:

h = b² / 2

If you go back and find

h = √(r² - k²)

then you'd find

k = a² / 2

 

[BvU]

Must be mistaken. One way to see that is from dimensions (this a physics community, after all ): left and right have different dimensions (length and length², respectively).

Back to the drawing board...

So where are the points that have the same distance from (0,0) and (b,0)

 

[Pratik89]

It was incorrect, those coordinates belong to the centroid

 

[LCKurtz]

An interesting property to note is that, the circumcenter (the center of the circle passing through the three points) has coordinates which are the average of all three. thus the center of the circle would be (b/3,a/3). This makes your job a lot easier

No, that is not where the circumcenter is. The points form a right triangle so the hypotenuse is a diameter. That is what makes the problem easier.

 

[Pratik89]

No, that is not where the circumcenter is. The points form a right triangle so the hypotenuse is a diameter. That is what makes the problem easier.[/QUOTE

No, that is not where the circumcenter is. The points form a right triangle so the hypotenuse is a diameter. That is what makes the problem easier.

Agreed

 

[ehild]

No, that is not where the circumcenter is. The points form a right triangle so the hypotenuse is a diameter. That is what makes the problem easier.

Making a sketch, one sees that it is really a right triangle and where the center of the circle is, and to find the radius is also very easy.

 

[BvU]

Pratik wants answers instead of questions, not guidance to help him find it by himself (and thus gain some insight). OK.
Well, the points that are equidistant from (0,0) and (b,0) are on x=b/2.
Once you know what you are looking for, a parametric equation isn't so hard to write down.
Erszebeths's picture is worth a thousand words ... or more.

And the culture at PF should remain as it is in this respect: learn a person to fish instead of giving him/her a fish.