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.Then locus of it's orthocenter is

• A. + = 100
• B. + = 100
• C. + = 100
• D. + = 100

Answer 1

The correct option is D + = 100

Distance of all the points from (0, 0) are 5 unit. That means circumcentre of the triangle formed by the given point is (0, 0). If G = (h, k) be the centroid of triangle, then
$3h=3+5(cos\theta +sin\theta ),3k=4+5(sin\theta -cos\theta )$
If $H(\alpha ,\beta )$ be the orthocenter, then
$OG:GH=1:2\Rightarrow \alpha =3h,\beta =3kcos\theta +sin\theta =\frac{\alpha -3}{5},sin\theta -cos\theta =\frac{\beta -4}{5}\Rightarrow sin\theta =\frac{\alpha +\beta -7}{10},cos\theta =\frac{\alpha -\beta +1}{10}Thus,locusof(\alpha ,\beta )is(x+y-7{)}^2+(x-y+1{)}^2=100$
Hence, (d) is the correct answer.