A line moves in the plane so that it passes through the point (1, 1) and such that it intersects the two coordinate axes. Find the locus of the centre of the circle which passes through these two points of intersection of the line with the coordinate axes, and through the origin.

i was thinking that :

enter image description here

the locus of the centre of the circle which passes through these two points of intersection of the line with the coordinate axes, and through the origin is $(x-1)^2 + (y-1)^2 =1^2 $

is its correct or not ? Pliz verified and give me some hints or any solutions

thanks in advance


Parametrization for circle coordinates would be is convenient.

$$ \frac{x}{a}+ \frac{y}{b} =1 $$

Since the straight line goes through $(1,1)$ we have

$$ \frac{1}{a}+ \frac{1}{b} =1 $$

Let $\alpha$ be the variable parameter of inclination of the straight line to negative x-axis


$$ c= \cos \alpha, s = \sin \alpha,\, s^2+c^2 =1;\, \frac{1}{a} = c^2 ;\, \frac{1}{b} =s^2 $$

and the parametric circles can be found from

$$ (x-a/2)^2 + (y-b/2)^2 = \frac{a^2+b^2}{4} $$

the above plugged in as

$$ (x-1/2c^2)^2 + (y-1/2s^2)^2 = \frac{(1/2c^2)^2 + (1/2s^2)^2}{4} $$

which you can take up further for simplification.


we have

$$ (x=1/2c^2) ; (y=1/2s^2); $$

Using relation $$ s^2+c^2=1 $$ we get $$ x+y = 2 x y ; \quad\frac1x + \frac1y =2\, $$ which relation of the hyperbola reminds one of the Lens formula in Optics. I cannot resist the temptation to add a sketch even if $OP$ did not ask for it.

$$ x\rightarrow u, \, y\rightarrow v, \, f\rightarrow \frac12; \quad \frac1u+\frac1v= \frac1f $$

enter image description here

If one of $(u,v)$ is given along with $f$ the other can be found out by such simple geometric construction suggested in OP's question with $(f,f)$ as concurrency point, put together with Newtonian Ray-trace construction.

It was not given in my high school physics text-book..

  • $\begingroup$ The OP was looking after the locus of the center of the circles. $\endgroup$ – Michael Hoppe Jan 16 '18 at 21:31
  • $\begingroup$ I left it as an exercise to OP./ $\endgroup$ – Narasimham Jan 16 '18 at 21:52
  • $\begingroup$ thanks a lots Narasimham $\endgroup$ – user469754 Feb 4 '18 at 9:12

The radius is not 1. By Pythagoras theorem, radius is $\sqrt{1^2+1^2}=\sqrt{2}$, found by finding the distance from the centre of the circle to the origin, which happens to be a point on the circumference.

  • $\begingroup$ im not not getting @Xcoderx $\endgroup$ – user469754 Jan 16 '18 at 11:11
  • $\begingroup$ @XcoderX pls note the correct latex for square root. $\endgroup$ – Deepak Jan 16 '18 at 11:12
  • $\begingroup$ We know that the formula of a circle right? Then the circle passes through the (0,0), and has (1,1) at the centre. So what is the distance between these 2 points? $\endgroup$ – QuIcKmAtHs Jan 16 '18 at 11:13
  • $\begingroup$ @Deepak thanks for the edit $\endgroup$ – QuIcKmAtHs Jan 16 '18 at 11:13
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    $\begingroup$ Yes that’s it! I hope you like my solution $\endgroup$ – QuIcKmAtHs Jan 16 '18 at 11:24

Any line through $(1,1)$ that has intersections with both axis is given by $y-1=m(x-1)$ and $m\neq0$. The intersection points are $(0,1-m)$ and $(1-1/m,0)$. These two points and the origin make a right triangle, so we are looking for Thales' circle. The center of the circle is the average of the intersection points, namely


Letting $x=1/2(1-1/m)$ gives $1/2(1-m)=\frac{x}{2x-1}$, hence the function $$x\mapsto y=\frac{x}{2x-1}$$ describes the locus of the center which answers the question. Notice that $1/x+1/y=2$

BONUS: The circle's radius is the distance of the center to the origin.

A little bit of algebra gives the equation of the circle: $$x^2-(1-1/m)x+y^2-(1-m)y=0.$$

Notice that this works for $m=1$, too.

CORIOUS REMARK: Pick a point of the $x$-axis and another of the $y$-axis. Then $y=\frac{x}{2x-1}$ cuts the line between them in two equals halves.

  • $\begingroup$ Michael.You are right, very much the same as yours, your answer more careful with m not=0.Anyway, one answer is enough , I deleted.Thanks for your comment. $\endgroup$ – Peter Szilas Jan 16 '18 at 22:47

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