# A triangle has vertices #A(a,b )#, #C(c ,d )#, and #O(0 ,0 )#. What is the equation and area of the triangle's circumscribed circle?

I generalized the question; let's see how that goes. I left one vertex at the origin, which makes it a bit less messy, and an arbitrary triangle is easily translated.

The triangle is of course totally inessential to this problem. The circumscribed circle is the circle through the three points, which happen to be the three vertices. The triangle does make a surprise appearance in the solution.

Some terminology: the circumscribed circle is called the triangle's circumcircle and its center the triangle's circumcenter .

Subtracting,

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For us that means

and a squared radius of

In Rational Trigonometry squared lengths are called quadrances and sixteen times the squared area is called the quadrea. We found the quadrance of the radius of the circumcircle is the product of the quadrances of the triangle divided by its quadrea.

If we just need the radius or area of the circumcircle, we can summarize the result here as:

The squared radius of the circumcircle is the product of the squared lengths of the triangle divided by sixteen times the triangle's squared area.

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To find the equation of the circumscribed circle of the triangle with vertices A(a,b), C(c,d), and O(0,0), first, we find the midpoint of each side of the triangle, which will also be the midpoint of the corresponding diameter of the circumscribed circle.

Let's denote the midpoints as M, N, and P for sides AC, AO, and CO respectively.

The coordinates of M, the midpoint of AC, can be found using the midpoint formula:

[ M = \left(\frac{a+c}{2}, \frac{b+d}{2}\right) ]

Similarly, the coordinates of N, the midpoint of AO, are:

[ N = \left(\frac{a+0}{2}, \frac{b+0}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) ]

And the coordinates of P, the midpoint of CO, are:

[ P = \left(\frac{c+0}{2}, \frac{d+0}{2}\right) = \left(\frac{c}{2}, \frac{d}{2}\right) ]

Now, we find the slopes of the perpendicular bisectors of the sides of the triangle, which will be the negative reciprocal of the slopes of the sides. Then, we can use the point-slope form to find the equations of these perpendicular bisectors.

Let ( m_{AC} ), ( m_{AO} ), and ( m_{CO} ) be the slopes of the sides AC, AO, and CO respectively.

[ m_{AC} = \frac{d-b}{c-a} ]

[ m_{AO} = \frac{b-0}{a-0} = \frac{b}{a} ]

[ m_{CO} = \frac{b-0}{c-0} = \frac{b}{c} ]

The negative reciprocals of these slopes are:

[ -\frac{1}{m_{AC}} = -\frac{c-a}{d-b} ]

[ -\frac{1}{m_{AO}} = -\frac{a}{b} ]

[ -\frac{1}{m_{CO}} = -\frac{c}{b} ]

Using the point-slope form with the midpoint coordinates, we get the equations of the perpendicular bisectors:

For side AC:

[ y - \frac{b+d}{2} = -\frac{c-a}{d-b} \left(x - \frac{a+c}{2}\right) ]

For side AO:

[ y - \frac{b}{2} = -\frac{a}{b} \left(x - \frac{a}{2}\right) ]

For side CO:

[ y - \frac{d}{2} = -\frac{c}{b} \left(x - \frac{c}{2}\right) ]

Solve the system of equations to find the point of intersection of these three perpendicular bisectors. This point will be the center of the circumscribed circle. Once you have the center point, you can find the radius by calculating the distance between the center and any of the vertices. The equation of the circumscribed circle will be in the form:

[ (x - h)^2 + (y - k)^2 = r^2 ]

where (h, k) is the center of the circle and r is the radius.

To find the area of the triangle's circumscribed circle, you would use the formula for the area of a circle:

[ A = \pi r^2 ]

where r is the radius of the circle.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- A circle has a chord that goes from #pi/3 # to #pi/2 # radians on the circle. If the area of the circle is #49 pi #, what is the length of the chord?
- A circle's center is at #(5 ,2 )# and it passes through #(2 ,3 )#. What is the length of an arc covering #(7pi ) /8 # radians on the circle?

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