What is the radius of a circle and its centre that passes through the points of 2 0 and 3 1 and -6 10 on the Cartesian plane showing work?

Answer 1

General equation of a circle on the Cartesian plane: x^2+y^2 +2gx+2fy+c = 0

Points: (2, 0) (3, 1) (-6, 10)

Substitute the above values into the general equation:-

4g+c = -4

6g+2f+c = -10

-12g+20f+c = -136

Solving the above simultaneous equations: g = 2, f = -5 and c = -12

General equation is now: x^2+y^2 +4x-10y-12 = 0

Completing the squares: (x+2)^2+(y-5)^2 -4-25-12 = 0

So: (x+2)^2+(y-5)^2 = 41

Therefore centre of circle is at: (-2, 5) and its radius is the square root of 41

Proof: Distance from (-2, 5) to any of the given points equals to the square root of 41

Answer 2

Let P = (2, 0), Q = (3,1) and R = (-6, 10)The gradient of PQ is 1 and its midpoint is (5/2, 1/2)

=> the perpendicular bisector of PQ, L1: (y - 1/2) = -1*(x - 5/2)

=> L1: x + y = 3.

The gradient of PR is 5/4 and its midpoint is (-2, 5)

=> the perpendicular bisector of PR, L2: (y - 5) = -4/5*(x + 2)

=> L2: 4x + 5y = 17

Then L2 - 4*L1 gives: 4x + 5y - 4x - 4y = 17 - 4*3

or y = 5 and then, since x + y = 3, x = -2

So the centre, C = (-2, 5).

Radius = sqrt(CP) = sqrt[(-2 - 2)^2 + (5 - 0)^2] = sqrt(1 + 25) = sqrt(41).