The base of triangle $ABC$ is $AB=6$ and the third vertex $C$ moves such that $\frac{sinA}{sinB}=2$. Then locus of vertex of $C$ is a circle then find the radius of circle.
Using sine rule we get $\frac{BC}{AC}=2$ but how to proceed further to get equation of cirlce?
Could someone help me with this?
Use coordinates . . .
$$\text{Let}\;A = (-3,0),\;\;B = (3,0),\;\;C=(x,y)$$
Then
\begin{align*} &\frac{\sin A}{\sin B}=2\\[8pt] \iff\;&\frac{BC}{AC}=2\\[8pt] \iff\;&\frac{BC^2}{AC^2}=4\\[8pt] \iff\;&\frac{(x-3)^2+y^2}{(x+3)^2 + y^2}=4\\[8pt] \end{align*}
Now simply cross-multiply, and express the resulting equation in the form $$(x-h)^2 + (y-k)^2 = r^2$$
Can you finish it?