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What is the locus of midpoints of the chords of contact of $ x^2+y^2=a^2$ from the points on the $\ell x + my + n = 0$

My approach is as follow

$\ell x + my + n = 0 \Rightarrow my = - \ell x - n \Rightarrow y = - \frac{{\ell x}}{m} - \frac{n}{m}$

Let us represent the tangent by $y = px + q \Rightarrow p = - \frac{\ell }{m};q = - \frac{n}{m}$

${x^2} + {y^2} = {a^2} \Rightarrow {x^2} + {\left( {px + q} \right)^2} = {a^2} \Rightarrow {x^2} + {p^2}{x^2} + {q^2} + 2xpq - {a^2} = 0$

$ \Rightarrow \left( {1 + {p^2}} \right){x^2} + 2xpq + {q^2} - {a^2} = 0 \Rightarrow {x^2} + \frac{{2xpq}}{{1 + {p^2}}} + \frac{{{q^2} - {a^2}}}{{1 + {p^2}}} = 0$

$h = - \frac{{pq}}{{1 + {p^2}}} = - \frac{{\frac{{\ell n}}{{{m^2}}}}}{{1 + \frac{{{\ell ^2}}}{{{m^2}}}}} = - \frac{{\ell n}}{{{m^2} + {\ell ^2}}}$, where $h$ represent the abscissa of the mid-point

${x^2} + {y^2} = {a^2} \Rightarrow {\left( {\frac{{y - q}}{p}} \right)^2} + {y^2} = {a^2}$

$ \Rightarrow {\left( {y - q} \right)^2} + {p^2}{y^2} = {p^2}{a^2} \Rightarrow {y^2} + {q^2} - 2qy + {p^2}{y^2} = {p^2}{a^2}$

$ \Rightarrow \left( {1 + {p^2}} \right){y^2} - 2qy + {q^2} - {p^2}{a^2} = 0 \Rightarrow {y^2} - \frac{{2qy}}{{1 + {p^2}}} + \frac{{{q^2} - {p^2}{a^2}}}{{1 + {p^2}}} = 0$

$k = \frac{q}{{1 + {p^2}}} = - \frac{{\frac{n}{m}}}{{1 + \frac{{{\ell ^2}}}{{{m^2}}}}} = - \frac{{mn}}{{{m^2} + {\ell ^2}}}\& h = - \frac{{\ell n}}{{{m^2} + {\ell ^2}}}$ where $k$ represent the ordinate of the mid point

$ \Rightarrow \frac{k}{n} = - \frac{m}{{{m^2} + {\ell ^2}}}\& \frac{h}{n} = - \frac{\ell }{{{m^2} + {\ell ^2}}}$

$ \Rightarrow \frac{k}{n} = - \frac{m}{{{m^2} + {\ell ^2}}}\& \frac{h}{n} = - \frac{\ell }{{{m^2} + {\ell ^2}}}$

$\frac{{{h^2}}}{{{n^2}}} + \frac{{{k^2}}}{{{n^2}}} = \frac{{{\ell ^2} + {m^2}}}{{{{\left( {{m^2} + {\ell ^2}} \right)}^2}}} \Rightarrow {h^2} + {k^2} = \frac{{{n^2}}}{{{m^2} + {\ell ^2}}}$

Not able to approach from here , the term $a^2$ is lost in calculation

Shubham Johri
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Samar Imam Zaidi
  • 9,202

1 Answers1

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If the tangents are drawn from the point $(\alpha,\beta)$ to the curve $S=0$ then the equation of chord of contact is $T=0\implies\alpha x+\beta y=a^2$

Also, if $(h,k)$ is the mid-point of the chord of contact then the equation of this chord is $T=S_1\implies hx+ky=h^2+k^2$

Comparing the two equations, we get $\dfrac{\alpha}h=\dfrac{\beta}k=\dfrac{a^2}{h^2+k^2}\implies\alpha=\dfrac{ha^2}{h^2+k^2}, \beta=\dfrac{ka^2}{h^2+k^2}$

Putting these values of $(\alpha,\beta)$ in $lx+my+n=0$, we get $\dfrac{hla^2}{h^2+k^2}+\dfrac{kma^2}{h^2+k^2}+n=0$

So, the required locus is: $la^2x+ma^2y+n(x^2+y^2)=0$

aarbee
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  • I am aware to this approach but I am trying to use different method of solving this problem. I am solving this question using the following method https://www.quora.com/What-is-the-locus-of-midpoints-of-the-chords-of-contact-of-x-2-y-2-a-2-from-the-points-on-the-lx-my-n-0#:~:text=The%20midpoint%20of%20chord%20of,chord%20of%20contact%20and%20line%20.&text=What%20is%20the%20locus%20of%20a%20point%20whose%20sum%20of,x%20%3D%202%20is%204%20units%3F&text=The%20slope%20of%20the%20line%20AB%20is%20%2D4%2F3. – Samar Imam Zaidi May 23 '21 at 08:01
  • @SamarImamZaidi That answer is wrong – Shubham Johri May 23 '21 at 08:15
  • @Shubham Johri I saw this post in quora and tried to solve it but I need to understand where I made mistake – Samar Imam Zaidi May 23 '21 at 08:26
  • @SamarImamZaidi The given line is not a tangent. You can't compare them. – aarbee May 23 '21 at 08:36