Question

Sol. [Key Concept: Two vertices on the circle $x^{2 +}y^{2 = 25}$ ⇒ orthocentre $=$ sum of the position vectors (circumcentre at $O$).]

By observation, it is clear that 3 vertices of triangle lie on circle $x^{2} + y^{2} = 25$.

$\therefore\ $ Centroid

$$G \equiv \left( \frac{5sin\theta + 5cos\theta + 3}{3},\frac{5sin\theta - 5cos\theta + 4}{3} \right) $$Circumcentre $O \equiv (0,0)$

⇒ Orthocentre, $H \equiv (h,k)$

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

$$\begin{array}{r} = 5sin\theta + 5cos\theta + 3\#(1) \end{array}$$

$$3\left( \frac{5sin\theta - 5cos\theta + 4}{3} \right) - 2(0) = k$$

$$\begin{array}{r} = 5sin\theta - 5cos\theta + 4\#(2) \end{array}$$

By eq. (1) + eq. (2),

$$h + k - 7 = 10sin\theta$$

By eq. (1) - eq. (2),

$$h - k + 1 = 10cos\theta$$

$\Rightarrow (x + y - 7)^{2} + (x - y + 1)^{2} = 100\ $ is the required locus.