Let the point P be the vertex of the parabola $y = x^{2} - 6x + 12$. If a line passing through the p

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Let the point P be the vertex of the parabola $y = x^{2} - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^{2} + y^{2} - 2x - 4y + 3 = 0$ at the points R and S , then the maximum value of $(PR + PS)^{2}$ is:

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Sol. [Key Concept: $(PR + PS)^{2}$ is largest when RS is a diameter, giving $(2\ PC)^{2}$.]Parabola $y = x^{2} - 6x + 12 = (x - 3)^{2} + 3y - 3 = (x - 3)^{2} \Rightarrow$ Vertex P $(3,3)$$${x^{2} + y^{2} - 2x - 4y + 3 = 0 }{C(1,2);r = \sqrt{2} }$$For max. of $(PR + PS)^{2}$RS should be diameter$$\begin{matrix} (PR + PS)_{\max}^{2} & \ = (PC - r + PC + r)^{2} \ & \ = (2PC)^{2} = 4(PC)^{2} \ & \ = 4 	imes 5 = 20 \end{matrix}$$

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