Solving a Challenging Calculus Math Problem - Can You Help?

[Wasper]

My Calculus math teacher often likes to give us math problems that we have never encountered , for home work , before he actually starts teaching it. According to him it gets us 'thinking' analyticaly . Anyways here's the problem that's gotten me stomped :

Find the equation of the circle passing through the points A (0,8) , B (0,6) , and C (0,0).

Here's what I've figured so far , points A and B forms a chord. I found the mid point of A and B which is (3,4). Then using the point of slop formula , I obtain a line for the diamater of the circle which is y= 3/4 X + 2.25 . At this point I can't figure out what to do next to obtain the radius or the center of the circle.

Some help please?

 

[recon]

Find the equation of the circle passing through the points A (0,8) , B (0,6) , and C (0,0).

Here's what I've figured so far , points A and B forms a chord. I found the mid point of A and B which is (3,4).

Some help please?

How can the midpoint of A and B be (3,4)? I think you made a typo for your question. There are no circles that can pass through (0,8), (0,6) and (0,0).

 

[Wasper]

Oops! Sometimes I type too fast and make silly errors . It's suppose to be (6,0) for point B NOT (0,6).

 

[Tom McCurdy]

ok
i got
center of the circle at 3,4
with a radius of 5

 

[Tom McCurdy]

It will all make 3,4,5 triangles in this fashion :)

 

[Tom McCurdy]

Algerbra way--the way i did the first time... silly me

Here is the algerbra way of doing it
General Form for circle equation
$$(x-x_0)^2+(y-y_0)^2=r^2$$

Here are the three versions of the equations

$$1. -x_0^2+(8-y_0)^2=r^2$$
$$2. (6-x_o)^2 + -y_0)^2 = r^2$$
$$3. -x_0^2+y_0^2=r^2$$

multiply equation three by negitive one
then add it to equation two to get
$$12x_0=36$$
therefore
$$x_0=3$$

again use the equation three and multiply it by negitive one
then add it to equation one to get
$$64=16y_0$$
therefore
$$y_0=4$$

then
simply
9+16=25
squareroot of 25
equals 5

 

[Tide]

Clearly the three points are the vertices of a right triangle and, therefore, the hypotenuse is a diameter (twice the radius!) of the circle so the midpoint of the hypotenuse is the center of the circle. The desired equation follows immediately.

 

[Wasper]

Ahhh thanks , but I've already solved it , well in a different way...

Using the point slope formula I got the equation for the diameter line Y=3/4X + 1 3/4. Then found the mid point between points AC and BC , then used point slop formula to get diameter line Y=4 for AC and X=3 for BC. I then simply prove that 3,4 is the center by setting y=4 and x=3 equal to y=3/4x +1 3/4 :

y=4,
4=3/4x + 1 3/4
-3/4x = 2.25
X = 3

x=3
y=3/4(3) + 1 3/4
y= 4

After that simply plug (3,4) and either point A , B , C back into the circle equation to get the radius.

Btw , how does the algebra way works? I don't understand it , does it involves something with triangles?