Problem 29 on the JEE mains inspired this note:

Things to know

•Pythagorean theorem

•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:

\( (y-a)^2+(x-b)^2=r^2\)

\(y^2+x^2=s^2\)

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.

## Comments

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TopNewestCan you tell me such way in finding common tangents to a circle &a ellipse ? – Anubhav Bhatia · 2 years, 3 months ago

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– Trevor Arashiro · 2 years, 3 months ago

That's why I made it a sub set of case 3. Because for all III, the second half of that inequality must be true.Log in to reply