finding the centre and radius of the circle that passes through A B and C?

Question

A (1,2) B (-3,-10) and C (-7,10)

How do you find the centre and the radius?

Is the perpenticular line for point B and C x= -5?

Answer 1

AB, AC and BC will be chords of the circle. The perpendicular bisector of each will pass through the centre of the circle. I will do AB and leave you to do either of the others. (You only need two to find the centre.)
Mid-point of AB has x coord. (1 + (-3))/2 = -1
and y coord. (2 + (-10))/2 = -4 so the mid-point is (-1,-4). The gradient of AB = (difference in y coords)/(difference in x coords) = (2 - (-10))/(1 - (-3)) = 12/4 = 3. Therefore the gradient of the perpendicular to this has gradient -1/3. Putting its gradient and the point it goes through together gives its equation
as y - (-4) = (-1/3)(x - (-1)) or y + 4 = (-1/3)(x + 1).
Find the equation of another perpendicular bisector in this way and then put them simultaneous to find where they intersect. You then find the radius by doing a pythagoras calculation on the centre and any one of the point A, B, C.

Answer 2

the respond above is a sturdy technique and could paintings. inspite of the undeniable fact that it supposes which you will remedy a device of three equations in 3 unknowns. attempt this for a greater "geometric" recommendations-set: Take any 2 of the standards, say A and B and discover the slope of the line between them. Take -one million over this slope as your new slope. additionally, discover the midpoint of AB. Use the slope and the midpoint to discover the equation of the line this is the perpendicular bisector of AB. Repeat this technique with yet another pair of things, say B and C. once you have finished, you have the equations for 2 strains that are perpendicular bisectors of aspects of triangle ABC. discover the intersection of those 2 strains (remedy 2 equations in 2 unknowns) and that's the middle O of your cirlce. ultimately, discover the area AO (or BO or CO) and you have the radius of circle O. this methodology works because you're looking the circumcircle of triangle ABC, that's the circle that passes interior the direction of the vertices of a triangle.

Answer 3

x² + y² + 2gx + 2fy + c = 0 is equation of circle

Point (1,2) gives:-
1 + 4 + 2g + 4f + c = 0
2g + 4f + c = - 5 -------Equation 1

Point (-3, - 10) gives:-
9 + 100 - 6g - 20f + c = - 109-----Equation 2

Point (- 7,10) gives:-
49 + 100 - 14g + 20f + c = 0----Equation 3

Subtract Equation 2 from Equation1:-
8g + 24f = 104------Equation 4

Subtract Equation 3 from Equation 2:-
8g - 40f = 40-------Equation 5

Subtract Equation 5 from Equation 4:-
64f = 64
f = 1
g = 10 (from Equation 5)

From Equation 1:-
20 + 4 + c = - 5
c = - 29

In general , equation of circle is:-
x² + y² + 2gx + 2 f y + c = 0
Centre (- g , - f)

In this question,
Centre (- 10,- 1) and c = - 29
radius = √(g² + f² - c) = √(100 + 1 + 29)
radius = √130