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A triangle has two of its sides along the co-ordinate axes and its third side is a tangent to the circle $x^2+y^2=a^2$. If the coordinates of the point of contact of the tangent are $(a \cosØ,a \sinØ)$, show that the coordinates of the centroid are $(a/3 \cosØ, a/3 \sinØ)$. Show that the locus of the centroid is $1/x^2+1/y^2=9/a^2$

Inceptio
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twa14
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1 Answers1

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From the Article $#148$ of this, the tangent at $(a\cos\phi,a\sin\phi)$ is $x\cos\phi+y\sin\phi=a$

So, $$\frac x{\frac a{\cos\phi}}+\frac y{\frac a{\sin\phi}}=1$$

So,the vertices are $(0,0);(\frac a{\cos\phi},0);(0,\frac a{\sin\phi})$

So, the centorid will be $$\left(\frac{0+0+\frac a{\cos\phi}}3,\frac{0+\frac a{\sin\phi}+0}3\right)$$

Now, eliminate $\phi$

lab bhattacharjee
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  • Thanks very much for your help, its a neat solution. I plotted out the locus in excel but I still have not managed to get the formula above 1/x^2+1/y^2=9/a^2. I go around in circles and end up with both sides of the equation the same!. Can you help me out here also? – twa14 May 07 '13 at 15:56
  • If $(\alpha,\beta)$ is the centorid, $\alpha=\frac a{3\cos\phi}\implies \cos\phi=\frac a{3\alpha}$

    Similarly, $\sin\phi=\frac b{3\beta}$

    $$\implies\left(\frac a{3\alpha}\right)^2+\left(\frac b{3\beta}\right)^2=1\implies \left(\frac 1{\alpha}\right)^2+\left(\frac 1{\beta}\right)^2=\left(\frac 3a\right)^2$$

    So, the locus of the cetroid $(\alpha,\beta)$ will be $$\frac 1{x^2}+\frac 1{y^2}=\frac9{a^2}$$

     – lab bhattacharjee May 07 '13 at 16:06
  • That's clear, many thanks again for your help – twa14 May 07 '13 at 21:26
  • Just an additional note, sinø= b/3β above should be sinø= a/3β. Using sin^2ϕ+cos^2ϕ=1 is again very neat. I was looking for relations between the distance from the origin to the centroid and to the IP with the tangent (a/2cosϕ,a/2sinϕ). It led nowhere. I would never have come up with your solution. Out of curiosity do you know of other ways of tackling this problem? – twa14 May 08 '13 at 08:12
  • If $(h,k)$ is the centorid and the equation of the tangent is $\frac xc+\frac yd=1,$ the vertices will be $(0,0);(c,0);(0,d)$

    So, $h=\frac{0+c+0}3\implies c=3h$. Similarly, $d=3k$

    So, the equation of the tangent becomes $\frac x{3h}+\frac y{3k}=1\iff k\cdot x+h\cdot y=3hk$

    Now, the distance of the center of the circle from the tangent is equal to the length of radius.

    So, $$\left|\frac{k\cdot 0+h\cdot 0-3hk}{\sqrt{k^2+h^2}}\right|=a$$

    Squaring we get, $$ 9h^2k^2=a^2(k^2+h^2)\iff \frac1{h^2}+\frac1{k^2}=\frac9{a^2}$$

     – lab bhattacharjee May 08 '13 at 08:37
  • Once again thanks for your input, I have learned a lot here. – twa14 May 08 '13 at 19:10