SOLUTION: A circle has x intercepts 0 and 4 and y intercepts 0 and 5 . Determine the equation of the circle. Can you please help me out? Thanks so much in advance:)

Algebra ->  Circles -> SOLUTION: A circle has x intercepts 0 and 4 and y intercepts 0 and 5 . Determine the equation of the circle. Can you please help me out? Thanks so much in advance:)      Log On


   

Question 825878: A circle has x intercepts 0 and 4 and y intercepts 0 and 5 . Determine the equation of the circle.
Can you please help me out? Thanks so much in advance:)
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--

THE PROBLEM:
A circle has x intercepts 0 and 4 and y intercepts 0 and 5 . Determine the equation of the circle.

A SOLUTION:
Since the circle has x-intercepts at 0 and 4 and y-intercepts at 0 and 5, the following
points are on the circle:
(0,0), (4,0), and (0,5)

The general form for the equation of a circle is %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2, where r is the
the radius and (h,k) is the center of the circle.

We need to find values for r, h, and k. Substitute our known values for (x,y) in to the general
form equation. Then we will have three equations and there variables.

I: For (x,y)=(0,0)
%28%280%29-h%29%5E2%2B%28%280%29-k%29%5E2=r%5E2
h%5E2%2Bk%5E2=r%5E2

Rewrite with k-expression on left:
k%5E2=r%5E2-h%5E2

II: For (x,y)=(4,0)
%28%284%29-h%29%5E2%2B%28%280%29-k%29%5E2=r%5E2
%284-h%29%5E2%2Bk%5E2=r%5E2

III: For (x,y)=(0,5)
%28%280%29-h%29%5E2%2B%28%285%29-k%29%5E2=r%5E2
h%5E2%2B%285-k%29%5E2=r%5E2

Rewrite with k-expression on left.
%285-k%29%5E2=r%5E2-h%5E2

Combine I and III (Both have r^2-h^2 on right-hand side.)
%285-k%29%5E2=k%5E2
k%5E2-10k%2B25=k%5E2

Solve for k. Subtract k^2 from both sides.
-10k%2B25=0
-10k=-25
k=2.5

Substitute 2.5 for k in equations I and II.
h%5E2%2Bk%5E2=r%5E2
h%5E2%2B%282.5%29%5E2=r%5E2
h%5E2%2B6.25=r%5E2

%284-h%29%5E2%2Bk%5E2=r%5E2
%284-h%29%5E2%2B%282.5%29%5E2=r%5E2
%284-h%29%5E2%2B6.25=r%5E2

Combine equations I and II. (Both have r^2 on the right-hand side.)
h%5E2%2B6.25=%284-h%29%5E2%2B6.25

Solve for h. Subtract 6.25 from both sides.
h%5E2=%284-h%29%5E2

Simplify.
h%5E2=h%5E2-8h%2B16

Subtract h^2 from both sides. Simplify.
0=-8h%2B16
8h=16
h=2

Substitute the unknown values (h,k)=(2,2.5) into the original equation for the circle.
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
%28x-2%29%5E2%2B%28y-2.5%29%5E2=r%5E2

Substitute the known value (x,y)=(0,0) into the equation and solve for r^2.
%280-2%29%5E2%2B%280-2.5%29%5E2=r%5E2
4%2B6.25=r%5E2
r%5E2=10.25

The equation for this circle is

10.25=10.25 CHECK!

%28%284%29-2%29%5E2%2B%28%280%29-2.5%29%5E2=10.25
10.25=10.25 CHECK!

%28%280%29-2%29%5E2%2B%28%285%29-2.5%29%5E2=10.25
10.25=10.25 CHECK!

Hope this helps! Feel free to email if you have any questions.

Mrs. Figgy
math.in.the.vortex@gmail.com