Math help?

Question

Write the standard form of the equation of the circle that passes through points at (1,1), (-2,2), and (5,1). Then identify the center and radius. Explain your solution.

Answer 1

You have three points, and we are told that they are on the circle. If you connect two of those points with a line, the bisector of that line will pass through the centre of the circle. So, find the bisector of two such segments:

(1, 1) to (-2, 2)
slope: -1/3
centre: (-1/2, 3/2)
bisector slope: 3
bisector: y - 3/2 = 3(x + 1/2)
y = 3x + 3

(1, 1) to (5, 1)
slope: 0
centre: (3, 1)
bisector slope: infinite
bisector: x = 3

Now, solve for the intersection of the two bisectors to find the centre of the circle:
x = 3
y = 3x + 3
y = 3*3 + 3
y = 12
centre: (3, 12)

Now find r², using the Pythagorean theorem with the centre and any one of the three given points:
r² = (3 - 1)² + (12 - 1)²
r² = 2² + 11²
r² = 4 + 121
r² = 125

Now, use the centre (3, 12) and the r² = 125 to find the standard form:

(x - 3)² + (y - 12)² = 125

Answer 2

Easiest way is to find the center. Do this by constructing perpendicular bisectors of lines connecting any two pairs of points, and solving the equations of the lines to get the intersection. One of these is particularly easy: x = 3. Once you have the center, you can easily get the radius, and then just plug the numbers into the standard equation.

Answer 3

The general equation of a circle is
(x - h)^2 + (y - k)^2 = r^2

You can rewrite this as
(x^2 - 2hx + h^2) + (y^2 - 2ky + k^2) = r^2
x^2 - 2hx + y^2 - 2ky = r^2 - h^2 - k^2
-2hx - 2ky - (r^2 - h^2 - k^2) = - x^2 - y^2
2hx + 2ky + (r^2 - h^2 - k^2) = x^2 + y^2
2hx + 2ky + c = x^2 + y^2, where c = (r^2 - h^2 - k^2)

Now plug in the three points to get 3 different equations:
2h + 2k + c = 2
-4h + 4k + c = 8
10h + 2k + c = 26

Solve this system for h, k and c. Use h, k, and the definition of c to find r.

Answer 4

using family of circles when 3 points on the circle are given the answer is(if i havent calculated wrong): x^2 + y^2 - 6x -24y + 28 = 0 and the center would be ( 3,12 ) and the radius would be 5*(5^1/2)....center is (-ve of half the coeff of x in the eqn of circle, -ve of half the coeff of y in the eqn of circle)....and the radius is square root of (sum of squares of coordinates of centers subtracting constant term of equation from the sum)....

Answer 5

their r many ways one wat is let the centre be h and k then u get three equ by finding the distance of the centre from these points and solving them becuase all have same distance equal to radius r

Answer 6

Here ya go:

http://www.regentsprep.org/Regents/math/geometry/GCG6/RCir.htm