Question

Given the base of a triangle and the ratio of the lengths of the other two unequal sides, prove that the vertex lies on a fixed circle.

Answer 1

BC length given

Assume $B$ $\to$ $(-a,o)$

$C$ $\to$ $(-a,o).$

$BC$= $2a$

$\frac{AB}{AC}=K(let,given)$

$\Rightarrow \frac{{AB}^2}{{AC}^2}=K^2$

$\Rightarrow$ $\frac{(x+a{)}^2+y^2}{(x-a{)}^2+y^2}=K^2.$

$\therefore$ We know K and we know $2a\left(ora\right).$

$\Rightarrow$ $(x+a{)}^2+y^2=k^2(x-a{)}^2+y^2{k}^2$

$\Rightarrow$ $x^2+a^2+2ax+y^2=k^2(x^2+a^2-2ax)+y^2{k}^2$

$\Rightarrow$ $x^2(k^2-1)+y^2(k^2-1)$ $-2ax(k^2+1)$$+(k^2{a}^2-a^2)$$=0.$

$\because$ coefficient of $x$${}^2$ = Coefficient of $y$${}^2$ and no coefficient of $xy$ term, hence the equation must be of circle.

$\therefore$ $(x,y)$ lie on circle.