Hi, I have the following question: find the standard equations of the circles that pass through (2,3) and are tangent to both the lines 3x -4y = -1 and 4x + 3y = 7.

My beginning of the solution: let the center of a circle be given by C(c,d).

Then I have the distance from (2,3) to C, given by

$\displaystyle \sqrt{(c-2)^2+(d-3)^2}$

This must be equal to the shortest distance to both the given lines, which is

$\displaystyle d = \frac{|3c - 4d + 1|}{5 } $

and also:

$\displaystyle d = \frac{|4c + 3d - 7|}{5 } $

Here I'm struck. I think this equation is not very helpful:

|3c - 4d + 1| = |4c + 3d - 7|

Can somebody give me a hint how to proceed? (I don't want the solution, I have it in my textbook)

Thanks in advance.