SOLUTION: Find the center and the radius of the circle that passes through the points (4,4) , (1,3) and (8,-4). What are the coordinates of the center? Thank you.

Algebra ->  Linear-equations -> SOLUTION: Find the center and the radius of the circle that passes through the points (4,4) , (1,3) and (8,-4). What are the coordinates of the center? Thank you.      Log On


   

Question 985218: Find the center and the radius of the circle that passes through the points (4,4) , (1,3) and (8,-4).
What are the coordinates of the center?
Thank you.
Found 3 solutions by MathLover1, Edwin McCravy, solver91311:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

same as Question 985229
Find the center and the radius of the circle that passes through the points
(4,4) , (1,3) and (8,-4).

equation of the circle is:
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 where h and k are x and y coordinates of the center, and r is radius
so use given points to set up system of three unknown
(4,4)
%284-h%29%5E2%2B%284-k%29%5E2=r%5E2....simplify
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1
(1,3)
%281-h%29%5E2%2B%283-k%29%5E2=r%5E2
1-2h%2Bh%5E2%2B9-6k%2Bk%5E2=r%5E2.............eq.2

(8,-4)
%288-h%29%5E2%2B%28-4-k%29%5E2=r%5E2....simplify
64-16h%2Bh%5E2%2B16%2B8k%2Bk%5E2=r%5E2.............eq.3
start with
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1
64-16h%2Bh%5E2%2B16%2B8k%2Bk%5E2=r%5E2.............eq.3
--------------------------------------------------------------------subtract
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2-%2864-16h%2Bh%5E2%2B16%2B8k%2Bk%5E2%29=r%5E2-r%5E2

-8h%2B16-8k-64%2B16h-8k=0
8h-16k=64-16.......both sides divide by 8
h-2k=8-2
h-2k=6...............eg.1a

go with
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1
1-2h%2Bh%5E2%2B9-6k%2Bk%5E2=r%5E2.............eq.2
---------------------------------------------------------subtract
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2-%281-2h%2Bh%5E2%2B9-6k%2Bk%5E2%29=r%5E2-r%5E2

22-8h-8k%2B2h%2B6k=0

-6h-2k=-22...............eg.2a
now go with
h-2k=6...............eg.1a
-6h-2k=-22...............eg.2a
-----------------------------------------------subtract
h-2k-%28-6h-2k%29=6-%28-22%29
h-cross%282k%29%2B6h%2Bcross%282k%29=6%2B22
h%2B6h=28
7h=28
h=28%2F7
highlight%28h=4%29
go to
h-2k=6...............eg.1a...plug in 4 for h
4-2k=6
4-6=2k
-2=2k
k=+-2%2F2
highlight%28k=+-1%29
go to
16-8h%2Bh%5E2%2B16-8k%2Bk%5E2=r%5E2.............eq.1 ..plug in 4 for h and -1 for k and find r
16-8%284%29%2B4%5E2%2B16-8%28-1%29%2B%28-1%29%5E2=r%5E2
cross%2816%29-cross%2832%29%2Bcross%2816%29%2B16%2B8%2B1=r%5E2
16%2B8%2B1=r%5E2
25=r%5E2
highlight%28r=5%29
so, your equation is: %28x-4%29%5E2%2B%28y%2B1%29%5E2=25

check:







Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Mathlover's way is rather difficult.  Substituting those points into 

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 and getting the system of three equations:



It's complicated but if you simplify all those and substitute, you'll
get h, k, and r, as she has shown above.

Here's an easier way but it's just about as long:  
Draw the three points and connect two pairs
of them.



We find the equations of the perpendicular bisectors of each chord.

The slope of the shorter chord, using the slope formula is 1/3

The midpoint of the shorter chord, using the midpoint formula is (5/2,7/2),

So the perpendicular bisector of the short chord has slope which is the negative
reciprocal of 1/3 which is -3, and it goes through (5/2,7/2)

So, using the point-slope equation of a line, and simplifying, the perpendicular
bisector of the shorter chord has equation y = -3x+11

Doing the exact same thing with the longer chord, we find that its slope is -2
and its midpoint is (6,0).  

So the perpendicular bisector of the longer chord has slope which is the
negative reciprocal of -2 which is 1/2, and it goes through (6,0).

Using the point-slope equation of a line, and simplifying, the perpendicular
bisector of the longer chord has equation y=expr%281%2F2%29x-3.

These perpendicular bisectors of the two chords are plotted in red below:



The two perpendicular bisectors (in red) must intersect at the center of the
circle, so we solve the system of equations:

system%28y+=+-3x%2B11%2C+y=expr%281%2F2%29x-3%29

and get their point of intersection as (4,-1), which is the center of the
circle.

We could use the distance formula to find the radius.  However it is not
necessary in this case because one of the given point (4,4), just happens
to be exactly 5 units above the center (4,-1), so we know that the radius
is 5.

So, since h=4,   k=-1, and  r=5, the standard equation   

%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 becomes

%28x-4%29%5E2%2B%28y%2B1%29%5E2=5%5E2

%28x-4%29%5E2%2B%28y%2B1%29%5E2=25.



Edwin

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


A perpendicular bisector of a chord of a circle passes through the center of the circle. Perpendicular bisectors of two different chords on the same circle will intersect in the center of the circle. The distance from the center to any of the endpoints of a chord is the radius of the circle.

1. Use the slope formula to calculate the slope of the line through one pair of your given points.

2. Use the midpoint formulas to calculate the coordinates of the midpoint of the line segment that joins the two points you selected for step 1.

3. Calculate the negative reciprocal of the slope you calculated in step 1.

4. Use the point-slope form of an equation of a line with the slope calculated in step 3 and the point determined in step 2 to derive the equation of the perpendicular bisector of the chord that joins the two points selected for step 1.

5. Repeat steps 1 through 4 for a different pair of given points.

6. The equations derived in steps 4 and 5 form a 2X2 linear system. Solve this system for the coordinates of the point of intersection. This will be the center of circle. For the moment, consider this point to be

7. Use the distance formula to determine the distance between and one of your given points. This will be , the radius of the circle.

8. Write the equation of the circle using the coordinates of the center and the value of the radius:



References

Slope Formula:



where and are the coordinates of the given points.

Slope relationship between perpendicular lines:



Midpoint formulas:

and



Point-slope form:



where are the coordinates of the given point and is the calculated slope.

Distance Formula:



.

John

My calculator said it, I believe it, that settles it