Question

Please explain the equation of the circle when the endpoints are (-3,4) and (6,1). (x-a)^2+(y-b)^2=r^2....Help I can not solve

Answer 1

I think the points they give you are on a diameter. The center of the circle is the point in between (-3,4) and (6,1).

can you find the mid point ?

Answer 2

the mid point will have the average x value and the average y value.
can you find the average of the x's -3 and 6
?

Answer 3

m=-3+6/2 ?

Answer 4

the x value of the center point (exactly between -3 and +6) is (-3+6)/2
you get 3/2 or 1.5

can you find the y value of the center point ? the average of the end point's y's

Answer 5

yes, I will try

Answer 6

(4+1)/2 I get 5/2 ?

Answer 7

looks good,

the center of the circle is (3/2 , 5/2)
so far we have
(x- 3/2 )^2 + (y- 5/2)^2 = r^2

we need the distance from the point (3/2 , 5/2) to one of the end points, for example (6,1)
do you know how to find the distance between 2 points ?

Answer 8

Yes

Answer 9

r^2 = (3/2 - 6)^2 + (5/2 - 1)^2

Answer 10

thank you so much for thanking the time to help me! :)

Answer 11

Can you find r^2 ?

Answer 12

Here is a graph of your circle

Answer 13

the r=22.5

Answer 14

r^2 is 22.5

(r would be sqrt(22.5)

the final equation is
(x- 3/2 )^2 + (y- 5/2)^2 = 22.5

Answer 15

You are so Awesome!! Thank you Much..

Answer 16

phi, I understand better because of you. once again take you.