Problem: Given any 3 non-collinear points, find the equation of the circle that contains them.
1. You need to select 3 non-collinear points, to use in the solution. The first point will be in quadrant 1, so choose a point with x and y coordinates between 5 and 25 first point:(___,___)
2. Let the 2nd point be in quadrant 2, again with values between 5 and 25, or -5 and -25, but select different values than you did for the first point
second point: (____,____)
3. Your 3rd point can be in quadrant 3 or 4. use the range of values as above, but again make sure the values are different from the first two points.
third point: (____,____)
4. Create a scatter plot
5. Location of mid-point between first and second point
x=_______ y=______
6. Location of mid-point between second and third point
x=_______ y=______
7.Equation of the perpendicular bisector of the line segment between the first and second point.
Y=_______
8.Equation of the perpendicular bisector of the line segment between the second and third point.
Y=_______
9. Center (___,___) rounded to 0.001
10. radius of the circle (rounded to 0.001)
r=_______
11. Equation of the circle
__________________
what, you don't know how to do any of those steps?
To find the equation of the circle that contains three non-collinear points, follow these steps:
1. Select three non-collinear points. Let's call them A, B, and C.
2. Determine the coordinates of each point. The first point should be in quadrant 1, with x and y coordinates between 5 and 25. Let's say A is (x1, y1).
3. Choose the second point (B) to be in quadrant 2, with values between 5 and 25 or -5 and -25. Make sure the values are different from the first point. Let's say B is (x2, y2).
4. The third point (C) can be in quadrant 3 or 4, using the same range of values as above. Again, make sure the values are different from the first two points. Let's say C is (x3, y3).
5. Create a scatter plot with the three points (A, B, C) on a coordinate plane.
6. Find the midpoint between points A and B. The x-coordinate of the midpoint is (x1 + x2) / 2, and the y-coordinate is (y1 + y2) / 2. Let's call this midpoint M1 = (xM1, yM1).
7. Find the midpoint between points B and C using the same formula. Let's call this midpoint M2 = (xM2, yM2).
8. Determine the equations of the perpendicular bisectors of the line segments AB and BC.
- The equation of the perpendicular bisector of AB can be found by using the midpoint M1 and the slope of the line segment AB. Use the formula y - yM1 = -1/m(AB)(x - xM1), where m(AB) is the slope of AB ((y2 - y1) / (x2 - x1)). Simplify the equation to obtain the equation of the perpendicular bisector of AB in the form y = mx + c1, where m is the slope and c1 is the y-intercept.
- Similarly, find the equation of the perpendicular bisector of BC using midpoint M2 and the slope of BC.
9. The intersection point of the two perpendicular bisectors is the center of the circle. Solve the system of equations formed by the two perpendicular bisector equations to find the coordinates of the center. Let's call the center point O = (xO, yO).
10. Calculate the distance between the center point O and any of the three given points (A, B, C). This distance is the radius of the circle. Use the distance formula: r = sqrt((x - xO)^2 + (y - yO)^2) where (x, y) is any of the three given points.
11. Finally, write the equation of the circle in the standard form: (x - xO)^2 + (y - yO)^2 = r^2. Replace xO, yO, and r with their calculated values to obtain the equation of the circle that contains the three given points.
Note: The steps outlined above can be performed using numerical calculations or graphical methods, depending on your preference and available tools.