Question

The base $AB$ of a triangle $ABC$ is fixed and vertex $C$ moves such that $\sin A=k\sin B$. If $AB=5$ then

• A. For $k\ne 1,$ locus of the vertex $C$ is a circle with radius of $\left|\frac{5k}{k^2-2k+1}\right|$
• B. For $k\ne 1,$ locus of the point $C$ is a circle with radius of $\left|\frac{5k}{1-k^2}\right|$
• C. For $k=1,$ the locus of the vertex $C$ is a line except the midpoint of $AB$
• D. For $k=1,$ the locus of the vertex $C$ passes through the midpoint of the line segment $AB$

Answer 1

Answer: (B), (C)

The correct options are
B For $k\ne 1,$ locus of the point $C$ is a circle with radius of $\left|\frac{5k}{1-k^2}\right|$
C For $k=1,$ the locus of the vertex $C$ is a line except the midpoint of $AB$


For $k=1$
$\sin A=\sin B\Rightarrow BC=AC$
Thus, the locus of the vertex $C$ is the perpendicular bisector of the line segment $AB$ except the midpoint of $AB$

For $k\ne 1$
$k=\frac{\sin A}{\sin B}=\frac{BC}{AC}$

$\Rightarrow B{C}^2=k^{2}A{C}^2\Rightarrow (5-x{)}^2+y^2=k^2(x^2+y^2)\Rightarrow 25+x^2+y^2-10x=k^2(x^2+y^2)\Rightarrow (1-k^2)x^2+(1-k^2)y^2-10x+25=0\Rightarrow x^2+y^2-\frac{10x}{1-k^2}+\frac{25}{1-k^2}=0$

So, the locus of the vertex $C$ is a circle with radius $\left|\frac{5k}{1-k^2}\right|$