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)?

Dec 4, 2016

Please see the explanation for steps leading to the equation: ${\left(x - 8\right)}^{2} + {\left(y - 1\right)}^{2} = {8}^{2}$

Explanation:

Given that the circle is tangent to the y axis at the point $\left(0 , 1\right)$, 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:

$3 x - 2 \left(1\right) - 22 = 0$

$3 x = 24$

$x = 8$

The center of the circle is $\left(8 , 1\right)$

The equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

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

${\left(0 - 8\right)}^{2} + {\left(1 - 1\right)}^{2} = {r}^{2}$

Solve for r:

$r = 8$

The equation of the circle is:

${\left(x - 8\right)}^{2} + {\left(y - 1\right)}^{2} = {8}^{2}$