How do you find the equation of a circl where.........?

Question

How do you find an equation of a circle where C(-4,2) and D(8,8) are endpoints of a diameter? How do you solve this Problem....

The answer is (x-2)^2+(x-5)^2=45...but HOW do you find this out???? Does the standard form of a circle have anything to do with this??? PLEASE HELp!!! THANKS

Answer 1

The equation of a circle is:
(x-h)^2 + (y-k)^2 = r^2
where:
h: x-coordinate of the center
k: y-coordinate of the center
r: Radius of the circle

The center of the circle will be midway between the points. On the x-axis:

(-4 + 8)/2 = 4/2 = 2

On the y-axis:

(2+8)/2 = 10/2 = 5

So the location of the center as coordinates is (2,5)
The radius is simply the distance from the center to either point. So, using the pathegreom theorm and the center at (h,k) and the point at (Px,Py):

sqrt((h-Px)^2 + (k-Py)^2) = r

Using (8,8) for the point and the center at (2,5):

sqrt((2-8)^2 + (5-8)^2) = sqrt((-6)^2 + (-3)^2) = sqrt(36 + 9) = sqrt(45)

Obviously, you could plug square root of 45 into your calculator to get some decimal equivilent, but it's not necessary since the equation requires r^2 anyways. So our final equation, using the form above, is:

(x-2)^2 + (y-5)^2 = (sqrt(45))^2

Simplifying, the square root and ^2 cancel out leaving:

(x-2)^2 + (y-5)^2 = 45

Answer 2

(x)^2 + (y)^2 = R^2 would be the standard form of a circle. R being the radius, of course.

When the circle center is at (a,b) the equation takes the form:
(x-a)^2 + (y-b)^2 = R^2 (notice (a,b) = (0,0) reduces to the standard form).

Now you know (8,8) and (-4,2) are points on the diameter, not just the circumference! Thus:

(8-a)^2 + (8 - b)^2 = R^2 = (-4-a)^2 + (2 - b) ^2

But note the (Euclidean) distance between (-4,2) and (8,8) gets you 2R (= the diameter). Ergo:

Distance = sqrt [ (8 - 2)^2 + (8 - (-4)) ^2]
= sqrt [ (36 + 144) ]
= sqrt [180]
= sqrt [45*4]
= 2sqrt[45]

==> R is sqrt(45). q.e.d.

Now fill in the R^2 back in the above equations,

(8-a)^2 + (8 - b)^2 = 45 = (-4-a)^2 + (2 - b) ^2

And you're done!

Answer 3

First find the distance between the endpoints and also find the midpoint of the endpoints

(-4,2) and (8,8)
D = sqrt((8 - (-4))^2 + (8 - 2)^2)
D = sqrt((8 + 4)^2 + 6^2)
D = sqrt(12^2 + 36)
D = sqrt(144 + 36)
D = sqrt(180)

Diameter = sqrt(180)
Diameter = 6sqrt(5)
Radius = (6sqrt(5))/2
R = 3sqrt(5) or sqrt(45)

m = ((8 + (-4))/2), ((8 + 2)/2)
m = ((8 - 4)/2),(100/2)
m = (4/2), 5
m = 2,5

Now use the formula

(x - h)^2 + (y - k)^2 = r^2

where as m = (h,k), and therefore h = 2, and k = 5

(x - 2)^2 + (y - 5)^2 = (sqrt(45))^2
(x - 2)^2 + (y - 5)^2 = 45

ANS : (x - 2)^2 + (y - 5)^2 = 45

Answer 4

Standard form of the circle is (x-a)^2+(y-b)^2=r^2
where r is the radius of the circle, and the centre of the circle is the point (a,b)

We can use Pythagoras' Theorem to find the length of the diameter, d say, namely
d^2=(8-(-4))^2+(8-2)^2
d^2=12^2+6^2=144+36=180

But d=2r (where r is the length of the radius)
so (2r)^2=180
ie 4r^2=180
r^2=45

Now we just need to find the centre of the circle. Clearly as the line CD forms a diameter, the centre of the circle lies halfway along the line CD.

So the centre has x-coordinate (-4+8)/2=2
and y-coordinate (2+8)/2=5

by averaging the x-coordinates of C and D, and then the y-coordinates of C and D.

So the centre of the circle is (2,5), and the radius squared is 45, so we have the equation (x-2)^2+(x-5)^2=45 using the standard equation of a circle.

Answer 5

Yes. To use the standard form of the circle, you need the coordinates of the center and the radius.

Find the midpoint of the line segment between C and D to find the center of the circle.
This will be at (2, 5)

Find the length of segment CD (it's the diameter, remember)
This will be sqrt(180) or 6 sqrt(5)

Divide it by 2 to determine the radius.
r = 3 sqrt(5)

Now you have all the information you need to plug into the standard form of the circle
(x -2) ^2 + (y-5) ^2 = [3sqrt(5)]^2 = 9x5 = 45

Answer 6

The center of the circle is halfway between C and D (because it is a diameter). Halfway between (-4, 2) and (8, 8) is (2, 5).

The length of the diameter is just the distance between C and D. Let E be (8, 2). Then CDE is a right triangle -- you should probably draw a diagram to make this clear. CE is 12, DE is 6, so CD (the diameter) is 6*sqrt(5). You want half that for the radius, which is 3*sqrt(5).

The equation for a circle whose center is (p, q) and whose radius is r is just:

(x-p)^2 + (y-q)^2 = r^2

Plugging in p=2, q=5 and r=3*sqrt(5), you get the equation you want, except you put (x-5)^2 when you want (y-5)^2.

Answer 7

You use the endpoints of the diameter to find the center of hte circle by using the midpoint formula. You should get (2, 5) as the center. Then you use the distance formula to find the radius. You should get sqrt of 45. Then you plug it into the standard equation of a circle, which you should get (x-2)^2 + (x-5)^2 = 45.
The standard equation of a circle is (x-a)^2 + (y-b)^2 = r^2, a&b being the points of the center.