Tangents are drawn to the circle $x^2+y^2=a^2$ from a point which always lies on the line $lx+my=1$. Prove that the locus of the mid point of chords of contact is $x^2 + y^2 -a^2(lx+my)=0$

My Attempt::

let mid point of chord of contact be $(h, k)$ & point $(x1, y1)$ be always lying on line: $lx+my=1$ then we have $lx_1+my_1=1$ ...(1) Now equation of chord of contact to circle: $x^2+y^2=a^2$ is $xx_1+yy_1=a^2$ but $(h, k)$ lies on chord of contact hence satisfying equation of chord gives us $hx1+ky1=a^2$ .......(2)

Please help me to complete this proof .

  • $\begingroup$ @@@Kanwaljit Singh, Could you please help me with this problem? $\endgroup$
    – pi-π
    Feb 1 '17 at 0:19
  • $\begingroup$ Observe that this midpoint is the circular inverse of the point on the line, so the resulting curve is the circular inversion of $lx+my=1$. $\endgroup$
    – amd
    Dec 16 '17 at 8:39

enter image description here

You can write the equation of the polar line (green line in the figure):


where $(x_1,y_1)$ is a point of the line $lx+my=1$ (red line).

The equation of the normal (blue line) to the polar line, passing through $(x_1,y_1)$ is:

$$y={y_1\over x_1}x$$

so we can find the middle point $M$ of the chord:

\begin{cases}x_1x+y_1y=a^2\\y={y_1\over x_1}x\end{cases}

Solving this system, we get:

$$M\left(x_M={a^2x_1\over x_1^2+y_1^2},y_M={a^2y_1\over x_1^2+y_1^2}\right)$$

Now we must eliminate $x_1$ and $y_1$, thus:

\begin{cases}x_M={a^2x_1\over x_1^2+y_1^2}\\y_M={a^2y_1\over x_1^2+y_1^2}\\lx_1+my_1=1\end{cases}

Squaring and adding the first and the second equation:

\begin{cases}x_M^2+y_M^2=a^4{x_1^2+y_1^2\over (x_1^2+y_1^2)^2}={a^4\over x_1^2+y_1^2}\\x_M={a^2x_1\over x_1^2+y_1^2}\\y_M={a^2y_1\over x_1^2+y_1^2}\\lx_1+my_1=1\end{cases}

\begin{cases}{x_M^2+y_M^2\over a^2}={a^2\over x_1^2+y_1^2}\\x_M={x_M^2+y_M^2\over a^2}x_1\longrightarrow a^2x_M=(x_M^2+y_M^2)x_1\\y_M={x_M^2+y_M^2\over a^2}y_1\longrightarrow y_M={x_M^2+y_M^2\over a^2}{(1-lx_1)\over m}\longrightarrow a^2my_M-(x_M^2+y_M^2)=-(x_M^2+y_M^2)lx_1\\y_1={1-lx_1\over m}\end{cases}

so $a^2my_M-(x_M^2+y_M^2)=-a^2lx_M\longrightarrow x_M^2+y_M^2-a^2(lx_M+my_M)=0$.

  • $\begingroup$ @@MattG88, Is this the continuation of my process? $\endgroup$
    – pi-π
    Feb 1 '17 at 0:41
  • $\begingroup$ Yes it could be;-) $\endgroup$
    – MattG88
    Feb 1 '17 at 0:42
  • $\begingroup$ @@MattG88, I could not understand clearly. Could you please say me what background knowledge is needed to tackle it? $\endgroup$
    – pi-π
    Feb 1 '17 at 0:46
  • $\begingroup$ I suppose that you need to know the equation of a line through a point, the equation of a polar line, the relation between the angular coefficients of two normal lines...I think it is enough;-) $\endgroup$
    – MattG88
    Feb 1 '17 at 1:01
  • $\begingroup$ However it also depends by the approach you want to use, it is possible to solve this problem applying the Euclid's theorem... $\endgroup$
    – MattG88
    Feb 1 '17 at 1:07

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