Question

what is the standard form of the equation of the circle with diametrical endpoints (5,-6) and (-3,2)?

Answer 1

what is the standard form of the equation of the circle with diametrical endpoints (5,-6) and (-3,2)?

Answer 2

TJ: Can you type in the form of the equation of a circle that has its center at (h,k) and has radius r?

Answer 3

by finding the midpoint, the center is (1,-2)
I need help finding the radius

Answer 4

Cool. I see you already know a lot, being able to recognize that you needed the coordinates of the center.
Keep in mind that the two given end points of a diameter of the circle, and the center of the circle, are on one and the same line. the line connecting your center (1,-2) to either of those points represents a radius of the circle; think for a moment, or draw a sketch, or use the pythagorean theorem, or simply divide the length of the diameter by 2, to obtain the radius.

Answer 5

How would you find the length of the diameter?
the length of the radius?

Answer 6

the distance formula

Answer 7

Right. You can apply the distance formula to find the radius, based on your having found the center of the circle. Are you comfortable with this? Or have you further questions about this problem?

Answer 8

I got 4root5 for the length of the diameter. Is this correct?

Answer 9

So, you've used the 2 points on the circle as the end points of the diameter, and are now trying to find the length of the diameter? OK: Here's how I'd do it:

Answer 10

\[diameter=\sqrt{((5-[-3])^{2}}+(-6-2)^{2})\]

Answer 11

does this resemble y our calculation?
Please evaluate the above expression yourself (all the numbers are actually under the radical).

Answer 12

\[\sqrt{(-3-5)^{2}+(2+6)^{2}]

Answer 13

Yes, and that boils down to \[\sqrt{(8^{2}}+8^{2}).\] Evaluate that, divide the result by 2, and you'll have your radius. OK?

Answer 14

ok thanks

Answer 15

You're very welcome. Hope to "see" you again on Open Study. Take care, TJ!