Indeterminate problem in geometry

[lomantak]

Hello,

I came across this question and am wondering if anyone could help me on this...I don't even know where to start... (see attached image)Any help would be greatly appreciated!

 

[mathman]

It is a somewhat tedious analytic geometry problem.

First locate Q by solving for the intersection of C1 equation and C2 equation - the latter is x^2+y^2=r^2.

By subtracting one equation from the other you will have a linear equation in x, which is readily solvable. This x can be used to obtain y (use plus value). Notice the coordinates of P and Q both depend on r.

Get the equation of the straight line through P and Q and find the x coordinate of the y=0 point on the line. This will also depend on r.

Now let r get arbitrarily small.

 

[lomantak]

Thank you, Mathman, for answering my question. I have done what you suggested by combining the equations of the two circles to find the intersection coordinates at Q.

C1: (x-2)^2 + y^2 = 4
C2: x^2 + y^2 = r^2

It comes out to
x = (r^2)/4

With this I can isolate r or x...either way, I now know the relation between the two variables.

But what I seem to have trouble with is the subtracting of the two equations. Do you mean subtracting C1 from C2's equation?
I find this odd, because it would then look like this:
x^2 + y^2 - 4x - (x^2 - 4x + 4 + y^2 -4) = 0
Then after everything, it just cancels out...so I know I must be plugging the x and r relation into the same equation... could you give me a little more guidance ... ,in detail, about this question?

You say there would then be a linear equation... but ... from where?Please overlook my ignorance.
Thanks!

 

[mathman]

I find this odd, because it would then look like this:
x^2 + y^2 - 4x - (x^2 - 4x + 4 + y^2 -4) = 0

No!

C1-C2 looks like: -4x+4=4-r^2. This gives the equation you used.