equation of a circle

Number1

Find the equation of a circle which is tangent to the x-axis at point (4, 0) and is tangent to line 3x-4y-17=0

Directed Distance, point slope formula...

Ok so I knew that the x-axis is the equation y=0. I think but i am not sure that the directed distance between the tangent line 3x-4y-17=0 is equal to the distance form (4,0) to the center. So this is what you get....

Abs(3x-4y-17)/Square root(9+4) = Square root ((x-4)^2+(y)^2)

But i am not sure how to get rid of the absolute value signs and i am not sure if this will get me anywhere. Will it get me to a dead end? If so, what do i do?

dacruick

it is tangent to the line 3x-4y-17=0 at what point? is it not given?

Mark44

You're assuming that the center of the circle is at (4, 0), which is not true. Assume that the center is at (4, k). You want the distance from the line to (4, k) to be equal to k, the radius of the circle.

If you square |x|, you get x².

Number1

I still don't understand. How do you know that the x value for the center is 4?? the x- value can be anything.

Zhivago

The distance from the center to point (4,0) must be the same as the distance from the center to the tangent line.

You know that the value of x coordiante for the center must be 4 because the circle is tangent to y=0 (the x-axis) at that point! Try to sketch it.


Knowing the slope of the tangent line let's you know the angle at which it touches the circle, and from here you can find the rest

Number1

Yea but if you try it out you get 2 circles!!! How do i know which one you are talking about? One is small another is large. The smaller one is on the bottom and the larger one is on the top.

Mark44

I haven't worked this problem, so it's possible that there are two circles. Can you be clearer when you say that one is "on the bottom" and one is "on the top"? On the bottom or top of what?

Number1

I mean on the top of the x axis or y=0 intercept and the bottom of the axis

Mark44

OK, that makes sense. The great big circle is tangent to the line at some point in the third quadrant, while the small circle is tangent in the first quadrant. Since the radii of the circles are different, you have two solutions. That's not a problem.

Mentallic

I didn't notice this till after drawing a diagram. The trick is that for the first big circle you'll have the equation in this form: [itex](x-4)^2+(y-k)^2=k^2[/itex] and then to find k you find the perpendicular distance from the line to the point (4,k).

For the second circle notice that while the radius is (say, r) it is in the 3rd quadrant so you'll have the circle equation in the form [itex](x-4)^2+(y+r)^2=r^2[/itex].
Notice how it is y+r because the r is going to be positive (since it's a distance) but the y coordinate of the centre of the circle is negative. Now you find the distance between the line and the point (4,-r).