SOLUTION: In an archaeological site, a piece of a broken plate was dug up. It was put on top of a grid, with the arc of the plate passing through A (7, 2), B (1, 4) and C (−7, 0). Find its

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: In an archaeological site, a piece of a broken plate was dug up. It was put on top of a grid, with the arc of the plate passing through A (7, 2), B (1, 4) and C (−7, 0). Find its      Log On


   



Question 1204888: In an archaeological site, a piece of a broken plate was dug up. It was put on top of a grid, with the arc of the plate passing through A (7, 2), B (1, 4) and C (−7, 0). Find its center and the standard equation of the circle describing the boundary of the plate.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the standard equation of the circle:
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

use point A (7, 2)
%287-h%29%5E2%2B%282-k%29%5E2=r%5E2
h%5E2+-+14h+%2B+k%5E2+-+4k+-+r%5E2+=+-53...solve for r
r+=+sqrt%28h%5E2+-+14h+%2B+k%5E2+-+4k+%2B+53%29....eq.1

use point B (1, 4)

%281-h%29%5E2%2B%284-k%29%5E2=r%5E2
h%5E2+-+2h+%2B+k%5E2+-+8k+-+r%5E2+=+-17
r+=+sqrt%28h%5E2+-+2h+%2B+k%5E2+-+8k+%2B+17%29....eq.2

and, use point C (-7,+0)

%28-7-h%29%5E2%2B%280-k%29%5E2=r%5E2
h%5E2+%2B+14h+%2B+k%5E2+-+r%5E2+=+-49
r+=+sqrt%28h%5E2+%2B+14h+%2B+k%5E2+%2B+49%29....eq.3

from eq.1 and eq.2 we have
sqrt%28h%5E2+-+14h+%2B+k%5E2+-+4k+%2B+53%29=sqrt%28h%5E2+-+2h+%2B+k%5E2+-+8k+%2B+17%29...square both sides
h%5E2+-+14h+%2B+k%5E2+-+4k+%2B+53=h%5E2+-+2h+%2B+k%5E2+-+8k+%2B+17
h%5E2+-+14h+%2B+k%5E2+-+4k+%2B+53-h%5E2+%2B2h+-+k%5E2+%2B+8k+-+17=0
-12h+%2B+4k+%2B+36+=+0...solve for k
k+=+3h+-+9......(1)

from eq.2 and eq.3 we have

sqrt%28h%5E2+-+2h+%2B+k%5E2+-+8k+%2B+17%29=+sqrt%28h%5E2+%2B+14h+%2B+k%5E2+%2B+49%29
h%5E2+-+2h+%2B+k%5E2+-+8k+%2B+17=+h%5E2+%2B+14h+%2B+k%5E2+%2B+49
h%5E2+-+2h+%2B+k%5E2+-+8k+%2B+17-+h%5E2+-+14h+-+k%5E2+-+49=0
-16h+-+8k+-+32+=+0
k+=+-2h+-+4.....(2)


from (1) and (2) we have
3h+-+9=-2h+-+4
3h+%2B2h=9+-+4
5h=5
h=1

go to
k+=+-2h+-+4.....(2), substitute h
k+=+-2+%2A1+-+4
k=-2-4
k=-6
=> the center is at (1,-6)

go to

r+=+sqrt%28h%5E2+%2B+14h+%2B+k%5E2+%2B+49%29....eq.3, substitute h+and+k
r+=+sqrt%281%5E2+%2B+14%2A1+%2B+%28-6%29%5E2+%2B+49%29
r+=+sqrt%28100%29
r+=+10

equation of your circle is:
x-1%29%5E2%2B%28y-%28-6%29%29%5E2=10%5E2
%28x-1%29%5E2%2B%28y%2B6%29%5E2=100