Shaelyn C.

asked • 03/06/18

circles equations

Find the equation of the circle that passes through the origin, has an x intercept of −8, and has a y intercept of 16. (The perpendicular bisector of a chord contains the center of the circle.)

Nathan A.

Greetings, Shaelyn.
 
We know three points that reside on the circle. Knowing the chord and thus the slope and midpoint between any two of them we can construct the perpendicular bisector.
 
We could use a bunch of calculations but roughly sketching what we know to make this easier.
Let us use (0, 0) and (-8, 0). The chord between them is horizontal thus the slope is undefined (vertical) for any perpendicular line. The mid-point is simply (-4, 0). We now have a point and slope for this perpendicular bisector, which gives us the linear equation, x = -4.
 
Now we repeat the process with any other combination of the three original points. Let us now use (0, 0) and (0, 16). This chord is vertical so any perpendicular line will be horizontal. The mid-point is (0, 8). This is gives us the line y = 8.
 
We now have two unique perpendicular bisectors that both contain the center of the circle, namely x = -4 and y = 8. The center of the circle is at their intersection and since the x and y values are respectively fixed the center is (-4, 8). This means for a standard circle equation we have h = -4, and k = 8.
 
All that remains is the radius, with is the distance from the center to any point on the circle, let us choose (0, 0). Then the radius = square root of [ (x2 - x1)^2 + (y2 - y1) ], which equals the square root of [ (-4 - 0)^2 + (8 - 0)^2 ] = square root of [80], we can stop there.
 
Substituting our values into the standard equation of a circle we have, (x - (-4))^2 + (y - 8)^2 = [square root of [80]]^2.
 
Simplifying, we end with (x + 4)^2 + (y - 8)^2 = 80
Report

03/06/18

1 Expert Answer

By:

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.