(a) The equation of the circle is
(x - 1)2 + (y - 2)2 = r2
Differentiating w.r.t to x, we get;
2(x - 1) + 2(y - 2)\(\frac{dy}{dx}\) = 0
⇒ (x - 1) + (y - 2)\(\frac{dy}{dx}\) = 0
(b) Given; cosx \(\frac{dy}{dx}\) + y = sinx; 0 ≤ x < \(\frac{π}{2}\)
⇒ \(\frac{dy}{dx}\) + sec xy = tanx
⇒ P = secx; Q = tanx