Expert: Scotto - 3/16/2010

Question
Write the equation of the circle that is tangent to the y-axis at (0,3) and contains the point (4,-1).

Ok, the thing i dont understand about this problem is how do they expect me to find the center of the circle?  There's just not enough information.  I tried using the midpoint formula and got that the point inbetween those two pts. is (2,1).  But i dont think that i can actually call that the center of my circle because the line connecting (0,3) and (4,-1) might just be a chord, not the diameter..  So I need help figuring out how to find out what the center of the circle is so i can write my equation.  I hope you can help me.

Answer
Tangent to the y axis means the edge of the circle barely touches the y axis at (0,3).

The equation of a circle is (x-a)² + (y-b)² = r².
What the problem gives is (4-a)² + (-1-b)² = r².

When it says it is tangent to the y-axis at (0,3), this means the center is at height 3.
In other words, b=3.

Putting this in the equation gives (x-a²) + (y-3)² = r².

When we know that (4,-1) is on the circle, that gives (4-a)² + (-1-3)² = r².
Since the circle is tangent at (0,3), we also know that (0,3) is on the circle,
so that gives (0-a)² + (3-3)² = r².

The 2nd equation comes down to a² = r², so a=±r.
Put this into the 1st equation to find r as a=r and a=-r.

See if both solutions work and make sense.
It may be that there are two possible circles.

When I think about it, there are two possible circles.  One circle has a large radius and the point on the same side of the circle as the axis it is tangent to and the other circle has a small radius and the point on the far side of the graph as the line.  When thinking of this, it is known that there are two solutions.