Question

How do you find the general form of the equation of a circle given the center at the point (-3, 1) and tangent to the y-axis?

Answer 1

Please see the explanation for the process. The equation of the circle is:

`(x - -3)^2 + (y - 1)^2 = 3^2`

Explanation:

The general equation for a circle is:

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

where `(h, k)` is the center point and `r` is the radius.

The center is `(-3, 1)`; this gives us the equation:

`(x - -3)^2 + (y - 1)^2 = r^2`

We are told that the y axis is tangent, therefore, the x coordinate is 0 and the y coordinate is the same as the center, 1. To find the value of r, temporarily substitute the point `(0,1)` for x and y in the equation:

`(0 - -3)^2 + (1 - 1)^2 = r^2`

`r = 3`

The equation of the circle is:

`(x - -3)^2 + (y - 1)^2 = 3^2`