Can you find the center of the circle that you can circumscribe about ΔEFG when E(2, 6), F(2, 2), and G(8, 2)?

Question

I'm probably going to need an explanation since I'll have to do other problems like this in the near future. Thanks in advance.

Answer 1

The center is the intersection of the perpendicular bisectors of the sides.

Of course you only need two of the perpendicular bisectors to find their intersection.

To see this, just imagine the side of the triangle as a chord of the circle. The center of the circle needs to be on the perpendicular bisector of the chord.

To get the perpendicular bisector (I mean the equation of the line for this), first find the midpoint of the side of the triangle (you have the two endpoints and you know the midpoint formula, right?). Now you have a point on the line, you just need the slope. But the slope is the negative reciprocal of the the slope of the side of the triangle, as they are perpendicular.

Now just use the point slope formula for a line (you know that one, right?)

Do this for another side. Then, solve the two linear equations simultaeously.

Answer 2

To circumscribe something, you need to find the point where the perpendicular bisectors of the sides meet. This will be the center of your circle. This will take a while, you've been warned ahead.

1. Pick two sides of the triangle, let's say EG and EF.
2. Find the midpoint of EG. I got (5,4).
3. Find the slope of EG. I got -2/3.
4. Take the opposite recriprocal of the slope of EG. This is the slope of the perpendicular bisector. I got 3/2.
5. With the midpoint of EG and the perpendicular slope, figure out the equation of the perpendicular bisector of EG. I got the equation as y =(3/2)x - (7/2)
6. Repeat steps 2 - 5 for EF. I got the final equation as y = 2.
7. Now use a system of equations for both lines to find x and y.
I got x = (11/3) an y = 2. So the center of your circle is (11/3, 2).

Hope that helps and good luck!

Answer 3

Let the circumcentre be the point (h, k)

Then the circumcircle is (x - h)² + (y - k)² = r²

with E, F and G lying on it

So for E (2, 6),
(2 - h)² + (6 - k)² = r²
ie h² - 4h + 4 + k² - 12k + 36 = r²

whence h² - 4h + k² - 12k = r² - 40 ... (1)

For F (2, 2),
(2 - h)² + (2 - k)² = r²
ie h² - 4h + 4 + k² - 4k + 4 = r²

whence h² - 4h + k² - 4k = r² - 8 ... (2)

And for G (8, 2),
(8 - h)² + (2 - k)² = r²
ie h² - 16h + 64 + k² - 4k + 4 = r²

whence h² - 16h + k² - 4k = r² - 68 ... (3)

ie h² - 4h + k² - 12k = r² - 40 ... (1)
h² - 4h + k² - 4k = r² - 8 ... (2)
h² - 16h + k² - 4k = r² - 68 ... (3)

Equation (2) - equation (1)
8k = 32
So k = 4

Equation 2 - equation (3)
12h = 60
So h = 5

r² = 13 so r = √(13)

Thus
the incentre is (5, 4) and the radius of the circumcircle is √(13)

Answer 4

EF and FG are two lines of which you know the endpoints. Take the perpendicular bisector of each of these lines. What is the equation of each of these lines? Now these two lines, the perpendicular bisectors, will meet at a point; that point is the center of the circle. The radius of the circle is the distance from the center to E,F, or G.

Answer 5

slope of EF=-4/0
slope of perpendicular=0/4
midpoint of EF=(2,4)
equation of the perpendicular bisector of EF
=y-4=0/4(x-2)
4y-16=0
y=4
slope of FG=0/6
slope of the perpendicular=-6/0
midpoint of FG=(5,2)
equation of the perpendicular bisector of FG
=y-2=-6/0(x-5)
0=-6x+30
6x=30
x=5
so the circumcentre of the circle is (5,4)
you can find the equation of the perpendicular bisector
of EG and see if this satisfies the equation as a check

Answer 6

I think u most unite the middle of each edge to the opposite corner of the triangle then use the compass to draw the circle with the center being where they meet and the radius being to the middle of any triangle edge

Answer 7

Circumcenter: (5,4)

Pbisector of EF:
mpoint is ((2+2)/2,(6+2)/2)= (2,4)
line y=4

Pbisector of FG:
mpoint is ((2+8)/2,(2+2)/2)=(5,2)
line x=5

Intersection of lines y=4 and x=5 is (5,4)