Question

Vertices of a variable triangle are $(3,4)$, $(5\cos \theta ,5\sin \theta )$ and $(5\sin \theta ,-5\cos \theta )$ , where $\theta \in R$ Locus of it's orthocentre is

• A. $x^2+y^2+6x+8y-25=0$
• B. $x^2+y^2-6x-8y+25=0$
• C. $x^2+y^2+6x-8y-25=0$
• D. $x^2+y^2-6x-8y-25=0$

Answer 1

The correct option is D $x^2+y^2-6x-8y-25=0$

Circumcentre of the triangle is $(0,0)$ and centroid is

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

$\because$ Centroid divides the join of orthocenter and circumcentre in the ratio $2:1$

$\therefore h=3+5\cos \theta +5\sin \theta$...........................(i)

$k=4+5\sin \theta -5\cos \theta$..............................(ii)

where $(h,k)$ represents the orthocentre

From (i) & (ii)

$\Rightarrow \sin \theta =\frac{h+k-7}{10}$.............................(iii)

and

$\Rightarrow \cos \theta =\frac{h+k-1}{10}$..............................(iv)

(iii)${}^2$+(iv)${}^2$

$\Rightarrow (h+k-7{)}^2+(h+k-1{)}^2=100$

$\Rightarrow h^2+k^2-6h-8k-25=0$

$\therefore$ Locus of Orthocentre is

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

Hence, option D is correct.