Question

One vertex of a triangle of given species is fixed, and another moves along the circumference of a fixed circle ; prove that the locus of the remaining vertex is a circle and find its radius.

Answer 1

Let $\Delta ABC$ be the given triangle.

Let $A$ be the fixed point. we take our axis such that $A$ lies on the origin.

Let $B$ be the point on the fixed circle.

Since $AB$ has has a constant length, the only possibility is that the fixed circle is centered at $A$ and has radius $=$ length of $AB$.

To find the locus of vertex $C$. we assume the following:

$AB=a$

$BC=b$

$\angle CBA=\beta$

$m_{AB}=\alpha$

Now, $\overrightarrow{AB}=a\cos \alpha i+a\sin \alpha j$

$\overrightarrow{BC}=-b\cos (\alpha +\beta )i-b\sin (\alpha +\beta )j$

Also, $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$

Thus $\overrightarrow{AC}=(a\cos \alpha -b\cos (\alpha +\beta \left)\right)i+(a\sin \alpha -b\sin (\alpha +\beta \left)\right)j$

Now, here do notice that $\beta$ is a fixed angle, whereas $\alpha$ is a variable.

Let us continue simplifying and grouping appropriate terms together to find that

$\overrightarrow{AC}=({\lambda }_1\cos \alpha +{\lambda }_2\sin \alpha )i+({\lambda }_1\sin \alpha -{\lambda }_2\cos \alpha )j$

Taking mod of both sides and squaring,

${\left|AC\right|}^2=a^2+b^2-2ab\cos \beta$

We see that the magnitude of $\overrightarrow{AC}$ is constant.

So, $C$ has a locus of circle with radius,

$r=\sqrt{a^2+b^2-2ab\cos \beta }$