Question

The base $AB$ of a triangle is fixed and its vertex $C$ moves such that $\sin A=k\sin B(k\ne 1)$. If $a$ is the length of the base $AB$, then the locus of $C$ is a circle whose radius is equal to

• A. $\frac{ak}{(2-k^2)}$
• B. $\frac{ak}{(1-k^2)}$
• C. $\frac{2ak}{(1-k^2)}$
• D. None of these

Answer 1

The correct option is B $\frac{ak}{(1-k^2)}$

Let the coordinates of $C$ be $(x_1,y_1)$ and the coordinates of $A$ and $B$ be $(a,0),$ respectively.

Given, $k=\frac{\sin A}{\sin B}=\frac{BC}{AC}$

$\Rightarrow {BC}^2=k^2{AC}^2$

$\Rightarrow {(x_1-a)}^2{x}_1^2+k^2(x_1^2+y_1^2)$

$\Rightarrow (1-k^2)x_1^2+(1-k^2)y_1^2-{2ax}_1+a^2=0$

$\Rightarrow x_1^2+y_1^2-\frac{2a}{1-k^2}x+\frac{a^2}{1-k^2}=0,[\because k\ne 1]$

Hence, locus of $C$ is $x^2+y^2-\frac{2a}{1-k^2}x+\frac{a^2}{1-k^2}-0,$

which is a circle whose centre is $(\frac{a}{1-k^2},0)$

and radius $\sqrt{\frac{a^2}{{(1-k^2)}^2}-\frac{a^2}{(1-k^2)}}=\frac{ak}{1-k^2}.$