circles.........HELP!!?

Question

ok so here's the killer---- find both the standard and general form equations for a circle with a diameter from (1,4) to (-5,-4)........

p.s i've calculated the center of this circle already...its (-2,0)
how do i write standard and general equations????
help!!

Answer 1

The equation of a circle is as follows:

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

(h,k) represents the coordinates of the center, and r represents the radius. Since you have the center, all you have to do is calculate the radius; the distance from the center to any one of those two points. Use the distance formula to get r.

Recall that the distance formula is
d = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 )

We want the distance from (-2,0) to (1,4), so
d = sqrt ( (1 - (-2))^2 + (4 - 0)^2 )
d = sqrt ( (3)^2 + 4^2 )
d = sqrt (25) = 5

Therefore, r = 5, and the equation of your circle is

(x - (-2))^2 + (y - 0)^2 = 5^2
(x + 2)^2 + y^2 = 25

Answer 2

You started out ok. The center (h,k) = (-2,0) as you said. Now you just need to calculate the radius r. It is the distance from the center to one of the points on the circumference. Let's pick the point (1,4).

r² = (-2-1)² + (0-4)² = 9 + 16 = 25
r = 5

So the standard equation of the circle is:

(x - h)² + (y - k)² = r²
(x + 2)² + (y - 0)² = 25

(x + 2)² + y² = 25

As for the general equation, perhaps they had this in mind:

(x + 2)² + y² = 25
(x + 2)² + y² - 25 = 0
x² + 4x + 4 + y² - 25 = 0

x² + y² + 4x - 21 = 0

This is the form of a general quadratic equation.

Answer 3

The center of the circle is at the midpoint of the
line joining the midpoint of the line joining your
2 given points. Averaging each coordinate,
we get (-2,0) as the center of the circle, as
you found.
Now we need the square of the radius
which is the distance² from (-2,0) to (1,4)
The distance² is (1+2)² + 4² =25.
So the equation of the circle is
(x+2)² + y² = 25.
To get the general equation of this circle,
expand this:
We get
x²+4x+y²-21 = 0.
Hope that helps!

Answer 4

In cartesian coordinates, the two forms of the circle equation are:
((x-xzero)/r)^2 + ((y-yzero)/r)^2 =1
and
(x-xzero)^2 + (y-yzero)^2 = r^2

Since xzero = -2 , yzero = 0 and r =3.
the two forms are:

((x+2)/3) ^2 + (y/3)^2 =1
and
(x+2)^2 + Y^2 = 9

The polar form is simpler. It is just r =3 with the center as you have stated.

Answer 5

The only equation that I can think of for your situation would be

(x - xc)^2 + (y-yc)^2 = r^2

where xc and yc are the x and y center points. standard and general I don't know without looking at your textbook.