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November 27, 2014

November 27, 2014

Posted by **Janice** on Friday, August 19, 2011 at 5:44pm.

So far I got

2D+6E+F+40

-4D+6E+F+52

-4D-2E+F-20

IS this correct> and how do I solve it?

- Math -
**MathMate**, Friday, August 19, 2011 at 6:33pmThe standard form of a circle centred at (x0,y0) with a radius of r is given by:

(x-x0)²+(y-y0)²=r² ...(1)

method 1: algebraic

Given the three points, we can substitute (x1,y1) in equation to obtain an equation in x0,y0 and r.

Substituting the other two points will give two more equations which should enable us to solve for x0, y0 and r.

method 2: geometric

Take any two of the three points as a chord of the circle. Find the perpendicular bisector of the segment, L1. Take the third point together with one of the two previous points and make another perpendicular bisector, L2.

The intersection of L1 and L2 will be the centre O (x0,y0) of the circle.

Find the radius r which is the distance of O to one of the three given points.

The unknowns of equation (1) are now all known.

method 3: by determinant

The equation of the circle is given by expanding the following determinant (columns unfortunately not lined up):

x²+y² x y 1

x1²+y1² x1 y1 1

x2²+y2² x2 y2 1

x3²+y2² x3 y3 1

Using method 3, I get

(x+1)²+(y-2)²=5²

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