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)?

1 Answer
Dec 4, 2016

Answer:

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#