Posted by **Kevin** on Thursday, December 15, 2011 at 12:27am.

using the knowledge of systems of equations, and that the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius. Find the equation of a circle that contains the following points: (6,2), (4,6), and (-3,5). (These 3 points each create an equation. Then you have a system. Use ELIMINATION to solve.)

- College Algebra -
**Reiny**, Thursday, December 15, 2011 at 10:40am
The method suggested for your question is not the best one in this case.

I will use the property that the centre lies on the intersection of the right - bisector of any two chords.

for (6,2) and (4,6)

slope = (6-2)/4-6) = -2

so slope of perpendicular is +1/2

midpoint of chord = ( (6+4)/2 , (2+6)/2 ) = (5,4)

equation of rt-bisector : y =(1/2)x + b

but (5,4) lies on it

4 = (1/2)(5) + b

b = 4-5/2 = 3/2

b = 3/2 --------> y = (1/2)x +3/2

for (6,2) and (-3,5)

slope of line = (5-2)/(-3-6) = -1/3

slope of perp is 3

midpoint of chord = (3/2, 7/2)

y = 3x + b

but (3/2,7/2) lies on it

7/2 = 3(3/2) + b

b = 7/2 - 9/2 = -1 ---- y = 3x - 1

intersection: 3x-1 = (1/2)x+3/2

times 2

6x - 2 = x + 3

5x = 5

x = 1

then y = 3(1) - 1 = 2

so we know the centre is (1,2)

using (6,2),

r^2 = (6-1)^2 + (2-2)^2

= 25

circle equation: (x-1)^2 + (y-2)^2 = 25

check:

for (6,2) --- LS = 25 + 0 = 25✔

for (4,6) --- LS = 9 + 16 = 25✔

for (-3,5) -- LS = 16 + 9 = 25✔

- College Algebra -
**Steve**, Thursday, December 15, 2011 at 10:47am
However, just to see how elimination works, plug in the points and solve the 3 equations. The trick is how to get rid of those pesky squared terms. Just equate the various values of r^2.

(6-h)^2 + (2-k)^2 = r^2

(4-h)^2 + (6-k)^2 = r^2

(-3-h)^2 + (5-k)^2 = r^2

(6-h)^2 + (2-k)^2 = (4-h)^2 + (6-k)^2

(6-h)^2 + (2-k)^2 = (-3-h)^2 + (5-k)^2

(4-h)^2 + (6-k)^2 = (-3-h)^2 + (5-k)^2

36-12h+h^2 + 4-4k+k^2 = 16-8h+h^2 + 36-12k+k^2

36-12h+h^2 + 4-4k+k^2 = 9+6h+h^2 + 25-10k+k^2

16-8h+h^2 + 36-12k+k^2 = 9+6h+h^2 + 25-10k+k^2

36-12h + 4-4k = 16-8h + 36-12k

36-12h + 4-4k = 9+6h + 25-10k

16-8h + 36-12k = 9+6h + 25-10k

-4h +8k = 12

-18h + 6k = -6

-14h - 2k = -18

Since we only have h and k, we only need two equations:

-h + 2k = 3

-6h + 2k = -2

5h = 5

h = 1

k = 2

(x-1)^2 + (y-2)^2 = r^2

pick any point:

(6-1)^2 + (2-2)^2 = 25 = r^2

r = 5

(x-1)^2 + (y-2)^2 = 25

## Answer This Question

## Related Questions

- College Algebra - using the knowledge of systems of equations, and that the ...
- NEED MAJOR HELP! *QUADRATIC EQUATIONS* - Write the equation for the circle with ...
- college algebra - write the equation of the circle in standard form find the ...
- College Algebra - Write the equation of a circle in standard Find the center, ...
- Algebra - 1.How do I find the center of a circle using the points (2,8) (-4,6) (...
- college pre-calculus - 4x^2+4y^2-16x-24y+51=0 is the equation of a circle. where...
- College Algebra - Write the equation of the circle in standard form. Find the ...
- MATH - How do i find the center of a circle using 3 points. Also need to find ...
- College Algebra - Write the equation of the circle in standard form. Find the ...
- college algebra - find the center and radius of the circle. Write in equation ...

More Related Questions