How do you write an equation for a circle with center (-8, -5) and tangent (touching at 1 point) to the y-axis?

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Answer 1 · 2016-11-25T20:18:11

Please see the explanation.

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

where $(x,y)$ is any point on the circle, $(h,k)$ is the center and r is the radius.

Substitute the given center into the equation:

$(x - -8)^2 + (y - -5)^2 = r^2$

Because the radius drawn to the point of tangency is always perpendicular, the point of tangency with the y axis must be $(0, -5)$

Substitute this point into the circle:

$(x - -8)^2 + (-5 - -5)^2 = r^2$

$r = 8$

The equation for the circle is:

$(x - -8)^2 + (y - -5)^2 = 8^2$