Problem 29 on the JEE mains inspired this note:
Things to know
•Graphical transformations (only vertical and horizontal shifts)
This note will describe a quick and easy way to find the number of common tangents to two circles.
Given the following equations for the circles:
find the number of common tangents these circles will have.
You may wonder, "this isn't generalized, you assume one is at the origin". Let me explain.
What I did was move both circles \(p\) units along the x-axis and \(q\) units along the y-axis to make one at the origin. The number of tangents to the circles won't change if we perform a transformation other than a distortion.
Now for the tangent rules: how many common tangents will they have if...
I) \(a^2+b^2=(r+s)^2\) then there are 3 common tangents.
II) \(a^2+b^2>(r+s)^2\) then there are 4 common tangents.
Now for case 3
III) \(a^2+b^2<(r+s)^2\) then we have three cases
NOTE: you must check this first, don't immediately check the rules below (Quick rule, if r=s in this case, then there are 2 tangents).
III-1) \(a^2+b^2=(r-s)^2\) then there is 1 common tangent.
III-2) \(a^2+b^2>(r-s)^2\) then there are 2 common tangents.
III-3) \(a^2+b^2<(r-s)^2\) then there are no common tangents.