What is the area of a circle in cm whose circumference passes through the points of 0 0 and 3 7 and 10 0 on the Cartesian plane showing work?

Answer 1

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

Points: (0, 0) (3, 7) (10, 0)

Substitute the points into the general equation to form simultaneous equations:-

0g+0f+c = 0

6g+14f+0c = -58

20g+0f+0c = -100

Solving the above simultaneous equations: g = -5, f = -2 and c = 0

Substituting the above values into the general equation: x^2 +y^2 -10x -4y +0 = 0

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

So: (x-5)^2 +(y-2)^2 = 29 which is the radius squared

Therefore area of the circle: 29*pi or about 91.11 square cm rounded to 2 d.p.

Answer 2

Let P = (0, 0), Q = (3, 7) and R = (10, 0) and let the centre, C, be at (u, v).C is on the perpendicular bisector of PR and so u = 5.

So PC^2 = (5 - 0)^2 + (v - 0)^2 = r^2 where r is the radius

25 + v^2 = r^2

Also QC^2 = (5 - 3)^2 + (v - 7)^2 = r^2

4 + (v - 7)^2 = r^2

25 + v^2 = 4 + (v - 7)^2

25 + v^2 = 4 + v^2 - 14v + 49

14v = 28 => v = 2

And then r^2 = 25 + v^2 = 25 + 4 = 29

So the equation is (x - 5)^2 + (y - 2)^2 = 29