Question

Help me find the equation of a circle with center (2,2); & passes thru the origin ?

Answer 1

Find the distance from (2,2) to (0,0). That will be the radius of the circle. Use the distance formula.

Answer 2

$$
r^2 = (x-h)^2 + (y-k)^2\qquad(\star)\\
$$

Where <h,k> is the center of the circle.

Now, you want the circle to also pass through the circle, so you want r to be the distance from the center of the circle to the origin:
$$
r = \sqrt{2^2 + 2^2}=\sqrt{2\times 4}=2\sqrt{2}
$$
So now you know, r and you also know h and k.
Plug these into the equation above (indicated by a \(\star\) above) to find the equation of a circle in terms of the variables x and y.

Answer 3

if it intersects axis at -1 and 3 does that mean I use the distance formula again ?

Answer 4

We don't care where it intersects the x axis. All we want is the equation of the circle.
We know the center is (2,2) and the radius is 2sqrt2 so radius squared is 8 and the equation is:
(x-2)^2+(x-2)^2=8 and that is the answer to the problem which says to find the equation of a circle.

Answer 5

No this is a new problem center is 1,2 and intersects x-axis at -1 and 3, does that mean I use the distance formula again using the -1 and 3 ?

Answer 6

Yes, that's what it means.

Answer 7

That means you need to find the radius by finding the distance from (1,2) to (3,0)

Answer 8

Oh okay, thank bro.

Answer 9

yw