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If `C_1,C_2,a n dC_3`belong to a family of circles through the points `(x_1,y_2)a n d(x_2, y_2)`prove that the ratio of the length of the tangents from any point on `C_1`to the circles `C_2a n dC_3`is constant. |
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Answer» Equations of circles `C_(1),C_(2)` and `C_(3)` through A`(x_(1),y_(1))` and `B(x_(2),y_(2))` are given by ` ubrace((x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2)))_("f(x,y)")` ` +lambda_(r)ubrace((y-y_(1)-(y_(2)-y_(1))/ (x_(2)-x_(2))(x-x_(1))))_("g(x,y)")=0,r=1,2,3` or `f(x,y)+lambda_(r)g(x,y)=0,r=1,2,3` Consider point P(h,k) on circle `C_(1)`. `:. f(h,k)+lambda_(1)g(h,k)=0` (1) Let `T_(2)` and `T_(3)` be the lengths of the tangents from P to `C_(2)` and `C_(3)`, respectively. `:. (T_(2))/(T_(3))=(sqrt(f(h,k)+lambda_(2)g(h,k)))/(sqrt(f(h,k)+lambda_(3)g(h,k)))` `=(sqrt(-lambda_(1)g(h,k)+lambda_(2)g(h,k)))/(sqrt(-lambda_(1)g(h,k)+lambda_(3)(h,k)))=sqrt((lambda_(2)-lambda_(1))/(lambda_(3)-lambda_(1)))` Clearly, this ratio is independtn of the choice of (h,k) and hence, a constant. |
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