Question

The vertices of a variable triangle are $(3,4),(5\cos \theta ,5\sin \theta ),$ and $(5\sin \theta ,-5\cos \theta ),$ where $\theta \in R.$ The locus of its orthocenter is

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

A] Given, vertices $(3,4)(5cos\theta ,5sin\theta )$

$(5sin\theta ,-5cos\theta )$ that lie on circle

$x^2+y^2=25$ where r = 5 , c =(0, 0)

circumcenter of $△$ is the origin.

centroid of $△$ is :-

G =$(\frac{5sin\theta +5cos\theta +3}{3},\frac{5sin\theta -5cos\theta +4}{3})$

as OG : GH = 1 : 3 [H is orthocenter]

$\therefore 3\left(\frac{5sin\theta +5cos\theta +3-2\times 0}{3}\right)=x$

$\Rightarrow x=5sin\theta +5cos\theta +3$

${111}^{y}y=5sin\theta -5cos\theta +4$

where $H=(x,y)$

Solving for $sin\theta$ & $cos\theta$

$sin\theta =\frac{x+y-7}{10}$ $cos\theta =\frac{x-y+1}{10}$

$\therefore$ locus is

$si{n}^2\theta +co{s}^2\theta =1$

$\Rightarrow (\frac{x+y-7}{10}{)}^2+(\frac{x-y+1}{10}{)}^2=1$

$\Rightarrow x^2+y^2-6x-8y-25=0$

Which is equation of circle.