Asked by Sally
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.
Answers
Answered by
MathMate
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)<sup>2</sup> + (y/yo)<sup>2</sup> = R<sup>2</sup>
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<sup>2</sup> + y<sup>2</sup> x y 1
x1<sup>2</sup> + y1sup>2</sup> x1 y1 1
x2<sup>2</sup> + y2sup>2</sup> x2 y2 1
x3<sup>2</sup> + y3sup>2</sup> x3 y3 1
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)<sup>2</sup> + (y/yo)<sup>2</sup> = R<sup>2</sup>
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<sup>2</sup> + y<sup>2</sup> x y 1
x1<sup>2</sup> + y1sup>2</sup> x1 y1 1
x2<sup>2</sup> + y2sup>2</sup> x2 y2 1
x3<sup>2</sup> + y3sup>2</sup> x3 y3 1
Answered by
MathMate
Sorry, the determinant does not show up very well, and there was a typo:
x<sup>2</sup> + y<sup>2</sup> x y 1
x1<sup>2</sup> + y1<sup>2</sup> x1 y1 1
x2<sup>2</sup> + y2<sup>2</sup> x2 y2 1
x3<sup>2</sup> + y3<sup>2</sup> x3 y3 1
x<sup>2</sup> + y<sup>2</sup> x y 1
x1<sup>2</sup> + y1<sup>2</sup> x1 y1 1
x2<sup>2</sup> + y2<sup>2</sup> x2 y2 1
x3<sup>2</sup> + y3<sup>2</sup> x3 y3 1
Answered by
MathMate
Please ignore the following format testing:
<pre>
x<sup>2</sup>+ y<sup>2</sup> x1 y1 1
x1<sup>2</sup>+y1<sup>2</sup> x1 y1 1
x2<sup>2</sup>+y2<sup>2</sup> x2 y2 1
x3<sup>2</sup>+y3<sup>2</sup> x3 y3 1
</pre>
<pre>
x<sup>2</sup>+ y<sup>2</sup> x1 y1 1
x1<sup>2</sup>+y1<sup>2</sup> x1 y1 1
x2<sup>2</sup>+y2<sup>2</sup> x2 y2 1
x3<sup>2</sup>+y3<sup>2</sup> x3 y3 1
</pre>
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