Question

How do you write the equation for a circle having center on the line 3x-2y-22=0 and tangent to the y-axis at (0,1)?

Answer 1

Please see the explanation for steps leading to the equation: `(x - 8)^2 + (y - 1)^2 = 8^2`

Explanation:

Given that the circle is tangent to the y axis at the point `(0,1)`, then the y coordinate of the center must be 1, because a radius drawn from the center to the point of tangency is perpendicular to the tangent line.

Substitute 1 for y in the given equation:

`3x - 2(1) - 22 = 0`

`3x = 24`

`x = 8`

The center of the circle is `(8,1)`

The equation of a circle is:

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

Substitute 0 for x, 8 for h, 1 for y, and 1 for k:

`(0 - 8)^2 + (1 - 1)^2 = r^2`

Solve for r:

`r = 8`

The equation of the circle is:

`(x - 8)^2 + (y - 1)^2 = 8^2`