Geo4.76

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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\,

Main Page:Geometry:Circles