You may see directly that perpendicular distance (distance along line passing through common centre) between both circumference will be difference of radii.

Still here is the solution using calculus...

Consider two circles \(C_{1}\) and \(C_{2}\) with common centre \((x_{0},y_{0})\) and radius \(r_{1}\) and \(r_{2}\) respectively

\(C_{1}:(x-x_{0})^{2}+(y-y_{0})^{2}=r_{1}^{2}\)

\(C_{2}:(x-x_{0})^{2}+(y-y_{0})^{2}=r_{2}^{2}\)

A general point on \(C_{1}\) can be assumed as \((x_{0}+r_{1}\cos\theta,y_{0}+r_{1}\sin\theta)\) and that on \(C_{2}\), it may be assumed as \((x_{0}+r_{2}\cos\theta,y_{0}+r_{2}\sin\theta)\).

Differentiating \(C_{1}\) wrt \(x\), we get
\(2(x-x_{0})+2(y-y_{0})y'=0\)\(\Rightarrow y'=-\dfrac{x-x_{0}}{y-y_{0}}=-\cot\theta\)

Similarly differentiating \(C_{2}\), we get \(y'=-\cot\theta\)

So tangents at points with same \(\theta\) are of same slope. Thus, we can say that they are parallel.
–
Pranjal Jain
·
2 years, 3 months ago

## Comments

Sort by:

TopNewestYes!

You may see directly that perpendicular distance (distance along line passing through common centre) between both circumference will be difference of radii.

Still here is the solution using calculus...

Consider two circles \(C_{1}\) and \(C_{2}\) with common centre \((x_{0},y_{0})\) and radius \(r_{1}\) and \(r_{2}\) respectively

\(C_{1}:(x-x_{0})^{2}+(y-y_{0})^{2}=r_{1}^{2}\)

\(C_{2}:(x-x_{0})^{2}+(y-y_{0})^{2}=r_{2}^{2}\)

A general point on \(C_{1}\) can be assumed as \((x_{0}+r_{1}\cos\theta,y_{0}+r_{1}\sin\theta)\) and that on \(C_{2}\), it may be assumed as \((x_{0}+r_{2}\cos\theta,y_{0}+r_{2}\sin\theta)\).

Differentiating \(C_{1}\) wrt \(x\), we get \(2(x-x_{0})+2(y-y_{0})y'=0\)\(\Rightarrow y'=-\dfrac{x-x_{0}}{y-y_{0}}=-\cot\theta\)

Similarly differentiating \(C_{2}\), we get \(y'=-\cot\theta\)

So tangents at points with same \(\theta\) are of same slope. Thus, we can say that they are parallel. – Pranjal Jain · 2 years, 3 months ago

Log in to reply