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I dont remember if I asked this question already but how do you find the equation of a circle that passes thru the origin and has intercepts equal to 1 and 2 on the x-axis and y-axis, respectively?
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- Feb 4, 2004
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In other words, you're looking for the circle with equation:
. . . . .(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>
...where (0, 0), (1, 0), and (0, 2) satisfy the equation? Just plug in the given points (which will give you three equations in h, k, and r), and try to solve the system.
Eliz.
. . . . .(x - h)<sup>2</sup> + (y - k)<sup>2</sup> = r<sup>2</sup>
...where (0, 0), (1, 0), and (0, 2) satisfy the equation? Just plug in the given points (which will give you three equations in h, k, and r), and try to solve the system.
Eliz.
Hello, Brit412!
You have a circle that passes through (1,0), (0,2), and (0,0).
. . We have an inscribed right triangle.
A right angle can be inscribed in a semicircle.
The center of the circle is the midpoint of the hypotenuse: (½, 1)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
The radius is half the length of the hypotenuse: √5/2
Equation: .(x - ½)<sup>2</sup> + (y - 1)<sup>2</sup> .= .5/4
Make a sketch!
You have a circle that passes through (1,0), (0,2), and (0,0).
. . We have an inscribed right triangle.
A right angle can be inscribed in a semicircle.
The center of the circle is the midpoint of the hypotenuse: (½, 1)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
The radius is half the length of the hypotenuse: √5/2
Equation: .(x - ½)<sup>2</sup> + (y - 1)<sup>2</sup> .= .5/4