Posted by **Sally** on Thursday, July 2, 2009 at 6:01pm.

Write the standard for of the equation of the circle that passes through the points at (0,8),(8,0),and (16,8). Then identify the center and radius of the circle.

- Math -
**MathMate**, Thursday, July 2, 2009 at 6:35pm
By inspection, the centre of the circle is at (8,8), since the three given points form a right-triangle in which the hypothenus is the diameter (0,8),(16,8).

The radius is half the diameter, namely (16-0)/2=8

The standard form of a circle with centre at (xo,yo) and radius of R is

(x/xo)^{2} + (y/yo)^{2} = R^{2}

Can you take it from here?

Note:

If you are looking for the general case of a circle passing through three distinct points P1(x1,y1), P2(x2,y2) and P3(x3,y3), it can be obtained by evaluating the following 4x4 determinant:

x^{2} + y^{2} x y 1

x1^{2} + y1sup>2 x1 y1 1

x2^{2} + y2sup>2 x2 y2 1

x3^{2} + y3sup>2 x3 y3 1

- Math -
**MathMate**, Thursday, July 2, 2009 at 6:37pm
Sorry, the determinant does not show up very well, and there was a typo:

x^{2} + y^{2} x y 1

x1^{2} + y1^{2} x1 y1 1

x2^{2} + y2^{2} x2 y2 1

x3^{2} + y3^{2} x3 y3 1

- Math -
**MathMate**, Thursday, July 2, 2009 at 9:42pm
Please ignore the following format testing:

<pre>

x^{2}+ y^{2} x1 y1 1

x1^{2}+y1^{2} x1 y1 1

x2^{2}+y2^{2} x2 y2 1

x3^{2}+y3^{2} x3 y3 1

</pre>

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