Let $P$ be a point on circumcircle of $\Delta ABC$, where $A=(3,4), B=(-3,4), C=(4,3)$. Let feet of perpendicular from $P$ to $AB$ and $AC$ be $Q$ and $R$, respectively. Then locus of circumcentre of $\Delta PQR$:
- Is a circle.
- Passes through origin.
- Passes through B and C.
- Passes through A.
It is quite obvious that circumcircle of $ABC$ is a circle centred at $O$ with a radius of $5$ units. To find circumcentre of $PQR$, I assumed $P$ as $(5\cos\alpha,5\sin\alpha)$ and found FoP from P to sides AB and AC. However, this is a lot of work, and finding the circumcentre of a triangle using a coordinate system is a tedious job. I am trying to use Euler's line in this case, but again it requires me to find the orthocentre. Is this the only way or am I missing something?
The simplicity of the answer is so amazing.
Okay, so consider $P$ as $(5cos\alpha,5sin\alpha)$, where $\alpha$ is a parameter.
After dropping perpendiculars on sides $AB$ and $AC$ (either can be produced), consider quadrilateral $PQAR$.
Since, $PQ \perp AQ$ and $PR \perp AR$, $\angle PQR + \angle PRA = 180^0$. Hence, $PQAR$ is a cyclic quadrilateral. Thus, a unique circle passes through points $P, Q$ and $R$ and this is the circumcircle of $\Delta PQR$.
Since $\angle PQA=90^0$, $AP$ is the diameter. Midpoint of $AP$ is the circumcentre $W(h,k)$.
$h=\frac{5cos\alpha+3}{2}$, $k=\frac{5sin\alpha+4}{2}$. Eliminate $\alpha$.
Locus of $W$ is $(h-\frac{3}{2})^2 + (k-2)^2 = \frac{25}{4}$.