Standard form of the equation for a circle?

Question

Okay, I was doing pretty well up until these...

1. Center at (3, 3); tangent to the x-axis

2. Center at (0, 0); passes through (2, 10)

3. Endpoints of a diameter are (5, 3) and (8, -3)

Someone PLEASE explain these to me!

Answer 1

Standard form - do you mean (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius?

If not skip the rest of this!

1. Since it's tangent to the x-axis its radius is 3 (the center is 3 above the x axis)
So (x-3)^2 + (y-3)^2 = 9

2. Find the radius squared using the distance formula
(2-0)^2 + (10-0)^2 = 4+100 = 104

So (x-0)^2 + (y-0)^2 = 104 or just x^2 + y^2 = 104

3. The center is the midpoint of the diameter ((5+8)/2, (3 + -3)/2) = (6.5, 0)

The radius is the distance from there to either diameter endpoint so again find the radius squared by the distance formula and plug in.
Hope that helps

Answer 2

The standard form of the equation for a circle is:

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

(h,k) represent the coordinates of the center of the circle
Hence, solving
1) (x - 3)^2 + (y - 3)^2 = r^2. Now, since the circle is tangent to the x-axis, it follows that it's radius, r, is equal to 3. Therefore,

(x - 3)^2 + (y - 3)^2 = 3^2 ANS

2) (x - 0)^2 + (y - 0)^2 = r^2.
Here, r = sqrt[2^2 + 10^2]
r = sqrt(104)

Therefore, x^2 + y^2 = [sqrt(104)]^2
x^2 + y^2 = 104 ANS

3) Solving for the radius first:
r = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
r = sqrt[(8 - 5)^2 + (-3 - 3)^2]
r = sqrt(45)

The center is at the midpoint of the line segment connecting (5,3) and (8,-3); hence,
xm = abscissa of midpoint = (x2 + x1)/2
xm = (5 + 8)/2 = 6.5
ym = ordinate of midpoint = (y2 + y1)/2
ym = 0

hence, (h, k) = (6.5, 0)

(x - 6.5)^2 + (y - 0)^2 = [sqrt(45)]^2
(x - 6.5)^2 + y^2 = 45 ANS


teddy boy

Answer 3

for number 3, the radius is the square root of 45 over 2

Don't forget; the radius is HALF the distance between the endpoints. So when you figure the distance using pythagorean theorem, you must divide that figure by half

so radius squared is 45 over 4 = 11.25

which makes the equation (x-6.5)^2 + y^2 = 11.25

(plug in the endpoints and you'll see it works)

Answer 4

(x-h)^2 +y-k)^2 = r^2 where (h,k) is center and r is radius

In number 1 , the radius is 3

In number 2 the radius is sqrt(10^2+2^2) = 2sqrt(26)

In number 3 center is at (6.5,0) and radius = sqrt(45)

That should give you all the data you need to plug values into standard equation and get your answers

Answer 5

I found this web site to be very helpful with circles.
http://cs.jsu.edu/mcis/faculty/leathrum/Mathlets/conics.html
It will show you how to line everything up.
(x-h)^2 = (y-k)^2 = r^2 is the standard form where (h,k) is the center and r is the radius