Math problem.

[needlotsofhelp]

the equation of a circle which contains the points A=(2,2) B=(3,-3) and c=(3,3)* is: x^2+y^2+_____x+_____y+12=0

I am lost, thanks for any help

 

[Guest]

The general form of the equation of a circle is
(x - a)^2 + (y - b)^2 = r^2

Using your 3 points A=(2,2) B=(3,-3) and C=(3,3) and also the centre D(a, b)

The distance
AD is SQRT[(b+2)^2 + (a+2)^2]
BD is SQRT[((b-3)^2 + (a+3)^2]
CD is SQRT[(b+3)^2 + (a+3)^2]

And the distance for each of these is the same, AD=BD=CD
then
[(b+2)^2 + (a+2)^2] = [((b-3)^2 + (a+3)^2]

b^2 + 4b + 4 + a^2 + 4a + 4 = b^2 -6b +9 + a^2 +6a +9

10b = 2a + 10

5b = a + 5 or a = 5b - 5

Repeat this process using BD = CD to get another equation. These can be solved to find (a,b) and then to write the circle equation.

 

[soroban]

Hello, needlotsofhelp!

A sketch simplied the problem dramatically . . .

The equation of a circle which contains the points A(2,2), B(3,-3), and C(3,3) is:
. . . . x² + y² + __x + __y + 12 = 0
.

Plot the three points.

You'll see that B and C are symmetric to the x-axis.
. . This means that the center lies on the x-axis.

The center lies on the perpendicular bisector of AC.
The slope of AC is 1. .The perpendicular slope is: m = -1.
The midpoint of AC is: (5/2, 5/2).

The line through (5/2, 5/2) with slope -1 is:
. . y - 5/2 .= .-1(x - 5/2) . ---> . y .= .-x + 5

This line intersects the x-axis when y = 0: .-x + 5 .= .0 . ---> . x = 5

Hence, the center of the circle is: P(5,0).

. . . . . . . . . . . . . . . . . . . . . . . . . . . .______________ . . . __
The radius is distance AP: .AP .= .√(5 - 2)² + (0 - 2)² .= .√13


The equation is: .(x - 5)² + y² .= .13

. . which simplfies to: .x² + y² - 10x + 0y + 12 .= .0