(8,-7) (6,3) what is its standar and general form?

Question

The center of the diameter of the circle are (8,-7) and (6,3)

Answer 1

In order to write the equation of this circle, we need two things, the center and the radius. Since the two points you were given are the endpoints of a diameter, the center is just the midpoint of those two points, and the radius is half of that diameter.

So lets start with the midpoint to find the center:

((8 + 6)/2, (-7 + 3)/2) = (7, -2)

Now we can use the distance formula to find out how long the diameter is:

sqrt((8 - 6)^2 + (-7 - 3)^2) = sqrt(4 + 100) = sqrt(104) = 2sqrt(26)

Remember that this is the whole diameter and not the radius, so we have to cut it in half. That gives us a radius of sqrt(26), which means r^2 = 26. This part is easy to miss, and using sqrt(104) as the radius would make the circle twice as big as it should be.

Now we can plug the center and r^2 into the standard form, which is (x - h)^2 + (y - k)^2 = r^2:

(x - 7)^2 + (y + 2)^2 = 26

To get the general form we just expand everything out and move it all over to one side:

x^2 - 14x + 49 + y^2 + 4y + 4 = 26

x^2 + y^2 - 14x + 4y + 27 = 0

So, our final answer is (x - 7)^2 + (y + 2)^2 = 26 for the standard form, and x^2 + y^2 - 14x + 4y + 27 = 0 for the general form.

Answer 2

THe radius is sqrt ( (8-6)^2 + (-7 -3)^2 )

= sqrt( 2^2 + (-10)^2 )

= sqrt( 4 + 100)

= sqrt(104)

If (8.-7) is the center then the equation is (x-8)^2 + (y+7)^2 = 104

if (6,3) is the center then the equation is (x-6)^2 + (y-3)^2 = 104