Question

If a point P has co-ordinates $(0,-2)$ and Q is any point on the circle, $x^2+y^2-5x-y+5=0$, then the maximum value of $(PQ{)}^2$ is.

• A. $8+5\sqrt{3}$
• B. $\frac{25+\sqrt{6}}{2}$
• C. $14+5\sqrt{3}$
• D. $\frac{47+10\sqrt{6}}{2}$

Answer 1

Answer: (C)

$14+5\sqrt{3}$

$(x-\frac{5}{2}{)}^2-\frac{25}{4}+(y-\frac{1}{2}{)}^2-\frac{1}{4}+5=0$

$=(x-\frac{5}{2}{)}^2+(y-\frac{1}{2}{)}^2=\frac{3}{2}$

On circle Q$=\frac{5}{2}+\sqrt{\frac{3}{2}}\cos Q,\frac{1}{2}+\sqrt{\frac{3}{2}}sinQ$

$P{Q}^2={\left(\frac{5}{2}+\sqrt{3/2}cosQ\right)}^2+{\left(\frac{5}{2}+\sqrt{3/2}\sin Q\right)}^2$

$P{Q}^2=\frac{25}{2}+\frac{3}{2}+5\sqrt{3/4}(\cos Q+\tan Q)$

$=14+5\sqrt{3/2}(\cos Q+\sin Q)$

$Ma{x}^{mr}=14+5\sqrt{3/2}\left(\sqrt{2}\right)$

$=14+5\sqrt{3}$.