Question

If the vertices of a variable triangle are $(4,3),(-5\cos \theta ,-5\sin \theta ),\text{and}(5\sin \theta ,-5\cos \theta )$, where $\theta \in R,$ then the locus of its orthocenter is

• A. $(x-y-1{)}^2+(x+y-7{)}^2=100$
• B. $(x+y-1{)}^2+(x-y-7{)}^2=100$
• C. $(x+y-7{)}^2+(x+y-1{)}^2=100$
• D. $(x+y-7{)}^2+(x-y+1{)}^2=100$

Answer 1

The correct option is A $(x-y-1{)}^2+(x+y-7{)}^2=100$

Clearly, all three points lie on $x^2+y^2=25.$
So, the circumcircle of the triangle formed by these points has centre at $(0,0).$
Let $H(h,k)$ be the orthocenter.

Now, we know that centroid divides the orthocenter and circumcircle in $2:1$ ratio
$\Rightarrow h=4-5\cos \theta +5\sin \theta ...\left(1\right),k=3-5\sin \theta -5\cos \theta ...\left(2\right)$

Adding and subtracting $\left(1\right)$ and $\left(2\right)$,
$h+k-7=-10\cos \theta$
$h-k-1=10\sin \theta$

Squaring and adding,
$(h-k-1{)}^2+(h+k-7{)}^2=100$

Hence, the locus will be
$(x-y-1{)}^2+(x+y-7{)}^2=100$