$P(a,b)$ is a point in the first quadrant.Circles are drawn through $P$ touching the coordinate axes,such that the length of the common chord of these circles is maximum.Find the ratio $a:b$.

The equation of the circle which touches both the coordinate axes is $x^2+y^2-2xr-2yr+r^2=0$.And as this circle passes through $P(a,b)$.

So $a^2+b^2-2ar-2br+r^2=0$
I am stuck here.I do not know what is the equation of length of the common chord of these circles.Please help me.

  • $\begingroup$ For each P, there is a single circle drawn through it, which is tangent to the two axes. Its center, obviously, lies on the main diagonal. $\endgroup$ – Lucian Dec 30 '15 at 18:46
  • $\begingroup$ @Lucian what you say is not true, see this $\endgroup$ – G-man Jan 2 '16 at 10:48
  • $\begingroup$ @G-man: True. There are two circles, depending on which of the two circular arcs point P lies. Thanks ! $\endgroup$ – Lucian Jan 2 '16 at 18:53

For each $P$ there will be two circles drawn through it that will touch both the axes and their centres will obviously lie on the line $y=x$ as shown:

enter image description here

As is evident from the figure, the common chord $PQ$ will always be the line segment joining the point $P$ to the point $Q$ which is the reflection of $P$ with respect to the line $y=x$ . Hence the coordinate of $Q\equiv(b,a)$ .

Now to maximise the length of the common chord, the chord $PQ$ should be a diameter of the circle passing through $P$ with it's centre at $\left(\dfrac{a+b}{2},\dfrac{a+b}{2}\right)$ and this circle must also touch both the axes. Can you find the answer now?

EDIT: The equation of the circle with PQ as its diamter is : $$(x-a)(x-b)+(y-a)(y-b)=0$$

If on solving this circle with the say, the X axis, if there are repeated roots, then it will touch said axis. so we get the condition: $$(a+b)^2=8ab$$

  • $\begingroup$ I am not sure how to maximise the length of the common chord(say $l$).$l^2=2(b-a)^2$.There are two variables $a$ and $b.$Can you give me more insight or hint?@G-man $\endgroup$ – Vinod Kumar Punia Jan 6 '16 at 16:22
  • $\begingroup$ @VinodKumarPunia check out the edit $\endgroup$ – G-man Jan 6 '16 at 16:34
  • $\begingroup$ $l^2=2(b-a)^2=2(a^2+b^2-2ab)$ and using $(a+b)^2=8ab$,we get $l^2=8ab$ which is maximum when $a=b$.So ratio $a:b$ is $1:1$.Am i right?@G-man $\endgroup$ – Vinod Kumar Punia Jan 6 '16 at 16:47
  • 1
    $\begingroup$ No, the constraint $(a+b)^2=8ab$ does NOT allow for $a=b$ . To find the ratio $a:b$, just divide this equation by $b^2$ and a quadratic in $a/b$ will result which is $x^2-6x+1=0$. There will be two soutions .which are $3\pm2\sqrt 2 $. @VinodKumarPunia $\endgroup$ – G-man Jan 6 '16 at 16:57

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