Vertices of a variable triangle are (3,4),(5 cos θ, 5 sin θ) and (5 sin θ,-5 cos θ), where θ ∈ R. Locus of it's orthocentre is

Question

Vertices of a variable triangle are (3,4),(5 cos θ, 5 sin θ) and (5 sin θ,-5 cos θ), where θ ∈ R. Locus of it's orthocentre 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

Tardigrade Admin · Accepted Answer

Clearly, circumcentre will be (6, 0) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/81274bc2ac2651468a89d9199a95a9fd-.png" /> (( h /3), ( k /3)) ≡((3+5 cos θ+5 sin θ/3), (4+5 sin θ-5 cos θ/3)) ⇒(( h -3/5))= cos θ+ sin θ (( k -4/5))- sin θ- cos θ cos θ=(( h - k +1/10)) sin θ=(( h + k -7/10)) ⇒( x + y -7)2+( x - y +1)2=100

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

$(x + y - 1)^{2} + (x - y - 7)^{2} = 100$

$(x + y - 7)^{2} + (x - y - 1)^{2} = 100$

$(x + y - 7)^{2} + (x + y - 1)^{2} = 100$

$(x + y - 7)^{2} + (x - y + 1)^{2} = 100$

Q. Vertices of a variable triangle are (3,4),(5cosθ,5sinθ) and (5sinθ,−5cosθ), where θ∈R. Locus of it's orthocentre is

(x+y−1)2+(x−y−7)2=100

(x+y−7)2+(x−y−1)2=100

(x+y−7)2+(x+y−1)2=100

(x+y−7)2+(x−y+1)2=100

Clearly, circumcentre will be (6,0) (3h​,3k​)≡(33+5cosθ+5sinθ​,34+5sinθ−5cosθ​) ⇒(5h−3​)=cosθ+sinθ (5k−4​)−sinθ−cosθ cosθ=(10h−k+1​) sinθ=(10h+k−7​) ⇒(x+y−7)2+(x−y+1)2=100

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

$(x + y - 1)^{2} + (x - y - 7)^{2} = 100$

$(x + y - 7)^{2} + (x - y - 1)^{2} = 100$

$(x + y - 7)^{2} + (x + y - 1)^{2} = 100$

$(x + y - 7)^{2} + (x - y + 1)^{2} = 100$