# If a point P has coordinates (0,-2) and Q is any point on the circle, x^2+y^2 -5x-y+5=0, then what is the maximum value of (PQ)^2?

## Mar 18, 2018

#### Explanation:

The center of the circle ${x}^{2} + {y}^{2} - 5 x - y + 5 = 0$ is $\left(\frac{5}{2} , \frac{1}{2}\right)$ and radius is $\sqrt{{\left(\frac{5}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2} - 5} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$

Let us put the values of the corrdinates of $P \left(0 , - 2\right)$ in the LHS of the equation of circle ${x}^{2} + {y}^{2} - 5 x - y + 5 = 0$ and we get

$0 + 4 - 0 + 10 + 5 = 19$ and as it is is greater than $0$, the pint is outside the circle and maximum distance of $P$ from any pont on the circle would be,

distance of $P$ from centrer of circle plus radius.

Distance of $P \left(0 , - 2\right)$ from center at $\left(\frac{5}{2} , \frac{1}{2}\right)$ is $\sqrt{{\left(\frac{5}{2} - 0\right)}^{2} + {\left(\frac{1}{2} - \left(- 2\right)\right)}^{2}} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \frac{5 \sqrt{2}}{2}$

and hence the maximum value of ${\left(P Q\right)}^{2}$ is

${\left(\frac{5 \sqrt{2}}{2} + \frac{\sqrt{6}}{2}\right)}^{2}$

= ${\left(5 \sqrt{2} + \sqrt{6}\right)}^{2} / 4$

= $\frac{50 + 6 + 20 \sqrt{3}}{4} = 14 + 5 \sqrt{3}$