Geo4.76

Find the locus of the poles of the line${\frac {x}{a}}+{\frac {y}{b}}=1\,$ w.r.t the circles which touch the coordinate axes and whose centre lies in the first quadrant.

Given line is ${\frac {x}{a}}+{\frac {y}{b}}=1\,$

or

$bx+ay=ab\,$

Equation of any circle which touches the coordinate axes and whose centre lies in the first quadrant is

$(x-\lambda )^{2}+(y-\lambda )^{2}=[\lambda ]^{2}\,$

$x^{2}+y^{2}-2\lambda x-2\lambda y+\lambda ^{2}=0\,$

Let P(x1,y1) be the pole of the line w.r.t the above circle, hence polar w.r.t. circle is

$xx_{1}+yy_{1}-\lambda (x+x_{1})-\lambda (y+y_{1})+\lambda ^{2}=0\,$

$(x_{1}-\lambda )x+(y_{1}-\lambda )y=\lambda (x_{1}+y_{1}-\lambda )\,$

Let this equation be 3.

since the first one and the 3 both represent polar of P w.r.t the circle

${\frac {x_{1}-\lambda }{b}}={\frac {y_{1}-\lambda }{a}}={\frac {\lambda (x_{1}+y_{1}-\lambda )}{ab}}\,$

From two of the above,

$ax_{1}-a\lambda =by_{1}-b\lambda \,$

$ax_{1}-by_{1}=\lambda (a-b)\,$

$\lambda ={\frac {ax_{1}-by_{1}}{a-b}}\,$

Taking the second two equalities,

$b(y_{1}-\lambda )=\lambda (x_{1}+y_{1}-\lambda )\,$

Substituting the value of lambda in this equation, we get

$b[y_{1}-{\frac {ax_{1}-by_{1}}{a-b}}]={\frac {ax_{1}-by_{1}}{a-b}}[x_{1}+y_{1}-{\frac {ax_{1}-by_{1}}{a-b}}]\,$

$b(ay_{1}-ax_{1})=(ax_{1}-by_{1})[{\frac {ay_{1}-bx_{1}}{a-b}}\,$

$(ax_{1}-by_{1})(ay_{1}-bx_{1})=b(a-b)(y_{1}-x_{1})\,$

Therefore, the required locus of P is

$(ax-by)(ay-bx)+b(a-b)(x-y)=0\,$