Q: Find the point P on the circle x^2 + y^2 – 4x – 6y +9 = 0 such that (i) ∠POX is minimum (ii) OP is maximum, where O is origin and OX is the x-axis.

Sol. Given circle is

x^{2} + y^{2} – 4x – 6y + 9 = 0 …. (i)

Its centre is C(2,3) and radius is 2.

(i) Let OP and ON be the two tangents from O to circle (i), then ∠POX will be minimum when OP is tangent to the circle at P. Let ∠POX = θ , $\displaystyle OP = \sqrt{(x^2 + y^2 -4x -6y +9)_{0,0}}$

= 3 = ON

∠ COP = (90-θ)/2 ,

$\displaystyle cos(\frac{90-\theta}{2})= \frac{OP}{OC} = \frac{3}{\sqrt{13}}$

$\displaystyle sin(\frac{90-\theta}{2})= \frac{2}{\sqrt{13}}$

sin (90 – θ) = cosθ = 12/13 and sinθ = 5/13

P ≡ (OP cosθ , OP sinθ) i.e. P≡ (36/13 , 15/13)

(ii) OP will be maximum , if P becomes the point where extended part of OC cuts the circle. Let this point be Q. Then maximum value of OP = OQ = OC + radius $=\sqrt{13} + 2 $

slope of line OC $\displaystyle = \frac{3-0}{2-0} = \frac{3}{2} = tan\alpha $ (say)

$\displaystyle cos\alpha = \frac{2}{\sqrt{13}} $ , $\displaystyle sin\alpha = \frac{3}{\sqrt{13}} $

For maximum OP,

P ≡ (OQ cosα , OQ sinα )

i.e. P≡ $\displaystyle 2+ \frac{4}{\sqrt{13}} , 3 + \frac{6}{\sqrt{13}}$