# Locus of circumcentre?

• Apr 30th 2009, 08:32 AM
fardeen_gen
Locus of circumcentre?
A triangle has two of its sides along the axes. Its 3rd side touches the circle $x^2 + y^2 - 2ax - 2ay + a^2 = 0$. Find the equation of locus of the circumcentre of the triangle.
• May 1st 2009, 12:37 AM
earboth
Quote:

Originally Posted by fardeen_gen
A triangle has two of its sides along the axes. Its 3rd side touches the circle $x^2 + y^2 - 2ax - 2ay + a^2 = 0$. Find the equation of locus of the circumcentre of the triangle.

I can't provide you with the equation of the locus but I can give you a few hints:

1.
$x^2 + y^2 - 2ax - 2ay + a^2 = 0~\implies~(x-a)^2+(y-a)^2=a^2$
The circle touches both coordinate axes.

2. You are dealing with a right triangle where the circumcentre is the midpoint of the hypotenuse.

3. If the tangentpoint is T(t, p) then the tangent has the equation:

$(x-a)(t-a)+(y-a)(p-a)=a^2$

4. If the tangent is parallel to the coordinate axes then there doesn't exist a circumcentre. Therefore the straight lines x = a and y = a must be asymptotes of the locus.

5. I've drawn a sketch: Triangles in blue, locus in red, asymptotes in green
• May 2nd 2009, 09:20 AM
pankaj
Let the circumcentre of the triangle $P(x_{1},y_{1}).$Therefore the the three vertices of the triangle will be $O(0,0),A(2x_{1},0)$ and $B(0,2y_{1}).$The Incentre of the triangle will be obviously $I(a,a)$ as can be seen in the diagram supplied by earboth.
Thus

$
a=\frac{(0)(2\sqrt{x_{1}^2+y_{1}^2})+(0)(2x_{1})+( 2x_{1})(2y_{1})}{2x_{1}+2y_{1}+2\sqrt{x_{1}^2+y_{1 }^2}}
$

Recall that the coordinates of the incentre of the triangle having vertices as $A(x_{1},y_{1}),B(x_{2},y_{2})$and $C(x_{3},y_{3})$is given by
$
(\frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+b y_{2}+cy_{3}}{a+b+c})
$

where $a=BC,b=CA,c=AB$

On simplification,we get

$
x_{1}+y_{1}+\sqrt{x_{1}^2+y_{1}^2}=\frac{2x_{1}y_{ 1}}{a}
$

Therefore the required locus is

$
x+y+\sqrt{x^2+y^2}=\frac{2xy}{a}
$

This is the equation of the curve which earboth has drawn in red.