How do you write the equation for a circle with points (0, 2a) (2b, 0) as ends of a diameter?

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How do you write the equation for a circle with points (0, 2a) (2b, 0) as ends of a diameter?

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Answer 1 · 2017-01-22T21:57:28

The x coordinate goes from 0 to 2b so the center is at b. Likewise, the y coordinate goes from 2a to 0 so the center is at a.
Use the equation ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}; h = b, k = a$ $(x-h)^2+(y-k)^2=r^2; h=b,k=a$ and one of the points.

Explanation:

Use the equation
${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

where $h = b and k = a$ $h=b and k=a$

${(x - b)}^{2} + {(y - a)}^{2} = r^{2}$ $(x - b)^2 + (y - a)^2 = r^2$

Let's use the point $(0, 2 a)$ $(0,2a)$ , to find the value of r:

${(0 - b)}^{2} + {(2 a - a)}^{2} = r^{2}$ $(0 - b)^2 + (2a - a)^2 = r^2$

$r^{2} = a^{2} + b^{2}$ $r^2 = a^2 + b^2$

$r = \sqrt{a^{2} + b^{2}}$ $r = sqrt(a^2 + b^2)$

Here is the equation of the circle:

${(x - b)}^{2} + {(y - a)}^{2} = {(\sqrt{a^{2} + b^{2}})}^{2}$ $(x - b)^2 + (y - a)^2 = (sqrt(a^2 + b^2))^2$

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How do you write the equation for a circle with points (0, 2a) (2b, 0) as ends of a diameter?

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