Answer:
(x-2)^2 + (y-3)^2 = 4
Step-by-step explanation:
Use the equation (x-h)^2 + (y-k)^2 = r^2 where the center of the circle is (h,k) and the radius is r.
We will see that all the points on the circle are of the form:
[tex](x, 3 \pm \sqrt{4 - (x - 2)^2} )[/tex]
A circle of radius R centered in (a, b) is defined by the equation:
(x - a)^2 + (y - b)^2 = R^2
In this case, the center is (2, 3) and the radius is 2, so we have:
(x - 2)^2 + (y - 3)^2 = 2^2
Now we want this in point form, so we need to isolate one variable.
(y - 3)^2 = 4 - (x - 2)^2
y^2 -2*3*y + 3^2 = 4 - (x - 2)^2
y^2 - 2*3*y + k = 0
Where k = 3^2 - 4 + (x - 2)^2
Then we can solve this with the Bhaskara's formula:
[tex]y = \frac{2*3 \pm \sqrt{(2*3)^2 - 4*1*(3^2 - 4 + (x - 2)^2)} }{2} \\\\y = 3 \pm \sqrt{+4 - (x - 2)^2}[/tex]
Then the points are of the form:
[tex](x, 3 \pm \sqrt{4 - (x - 2)^2} )[/tex]
If you want to learn more about circles, you can read:
https://brainly.com/question/1559324
