Vertices of a variable triangle are $A(3,4),B(5\cos \theta ,5\sin \theta )\text{ and }C(5\sin \theta ,-5\cos \theta )$ Then, locus of its orihocenire is

detailed solution

Correct option is D

We observe that OA =OB =OC. Therefore 0 (0, 0) is the circumcentre of△ABC. Let O′(h,k)  be the coordinates of its orthocentreThe centroid G 3+5cos⁡θ+5sin⁡θ3,4+5sin⁡θ−5cos⁡θ3divides the line segment joining OO′ in the ratio 1 : 2.∴ 1×h+2×01+2=3+5cos⁡θ+5sin⁡θ3and 1×k+2×01+2=4+5sin⁡θ−5cos⁡θ3O(0,0)1⟶ 2  O   O' (h,k)⇒ h=3+5cos⁡θ+5sin⁡θ and k=4+5sin⁡θ−5cos⁡θ⇒ h+k−7=10sin⁡θ and h−k+1=10cos⁡θ⇒ (h+k−7)2+(h−k+1)2=100Hence, the locus of (h,k) is (x+y−7)2+(x−y+1)2=100