It's actually my textbook problem so I have seen a proof in my class but that doesn't satisfy me. The proof was given by the condition that if $px+qy=1$ is a tangent then the distance from the center $(0,0)$ is equal to radius $a$. From there we find an equation like $p^2 + q^2 = 1/a^2$. As it is an equation of a circle, $(p, q)$ is situated on a circle. Yes I get the theoretical proof but I don't understand how it happens in real, in a cartesian plane. Please explain me how it would look like in a graph paper. What does the question even mean?

  • $\begingroup$ What they mean is that the locus, i.e the path traced out by the points p and q when they are under the restriction that $$p^2+q^2=\frac{1}{a^2}$$ is a circle $\endgroup$
    – Nikunj
    Jan 30, 2016 at 14:40

1 Answer 1


Let $(x',y')$ be the point of contact, then the equation of the tangent $L$ is $x'x+y'y=a^{2}$.

$L$ : $\displaystyle \frac{x'x}{a^{2}}+\frac{y'y}{a^{2}}=1$

$\displaystyle \therefore (p,q)= \left( \frac{x'}{a^{2}},\frac{y'}{a^{2}} \right)$

Since $(x',y')$ lies on the circle, $x'^{2}+y'^{2}=a^{2}$

$\displaystyle \implies p^{2}+q^{2}=\frac{x'^{2}}{a^{4}}+\frac{y'^{2}}{a^{4}}=\frac{a^{2}}{a^{4}}$

i.e. $(p,q)$ lies on the circle $\displaystyle x^{2}+y^{2}=\frac{1}{a^{2}}$

  • $\begingroup$ shouldn't it be p^2 + q^2 = 1/a^2? $\endgroup$ Feb 3, 2016 at 11:17
  • $\begingroup$ @FarhanFuad Yep, I forgot to square the $a^{2}$. Corrected and many thanks. $\endgroup$ Feb 3, 2016 at 15:21

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