# What is the max value?

Apr 9, 2018

The answer is (2)":- "color(red)(14+5sqrt3

Check in image...

#### Explanation:

Put the coordinate $\textcolor{red}{\left(0 , - 2\right)}$ on the equation and you will get the value greater than color(red)(0. It means the point color(red)(p(0,-2) is outside the circle.

Now check...

We determined the value of $\overline{\text{PQ}}$ by distance formula.

Not for JEE level, but a huge calculation. Will try to solve it with shorter method.

For any problem about handwriting , notify me.

:-)

Hope it helps...
Thank you...

Apr 9, 2018

$P {Q}^{2} = 14 + 5 \sqrt{3}$

#### Explanation:

given a circle : ${x}^{2} + {y}^{2} - 5 x - y + 5 = 0$,
rewrite the equation :
$\implies {\left(x - \frac{5}{2}\right)}^{2} + {\left(y - \frac{1}{2}\right)}^{2} - \frac{25}{4} - \frac{1}{4} + 5 = 0$
$\implies {\left(x - \frac{5}{2}\right)}^{2} + {\left(y - \frac{1}{2}\right)}^{2} = \frac{3}{2}$
let $C \left(x , y\right) \mathmr{and} r$ be the center and the radius of the circle, respectively.
$\implies C \left(x , y\right) = C \left(\frac{5}{2} , \frac{1}{2}\right) , \mathmr{and} r = \sqrt{\frac{3}{2}}$
maximum $P Q = P C + C Q = P C + r$
$P Q = \sqrt{{\left(\frac{5}{2} - 0\right)}^{2} + {\left(\frac{1}{2} + 2\right)}^{2}} + \sqrt{\frac{3}{2}}$
$= \sqrt{\frac{50}{4}} + \sqrt{\frac{3}{2}}$
$= \frac{5 \sqrt{2}}{2} + \sqrt{\frac{3}{2}}$
$= \frac{5 + \sqrt{3}}{\sqrt{2}}$
$\implies P {Q}^{2} = {\left(\frac{5 + \sqrt{3}}{\sqrt{2}}\right)}^{2} = 14 + 5 \sqrt{3}$

Solution 2:)

$\tan a = \frac{\frac{5}{2}}{\frac{1}{2} + 2} = 1 , \implies a = {45}^{\circ}$
$Q \left({x}_{q} , {y}_{q}\right) = \left(x + r \cos 45 + y + r \sin 45\right)$
$= \left(\frac{5}{2} + \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2} , \text{ } \frac{1}{2} + \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2}\right)$
=(5/2+sqrt3/2, ' " 1/2+sqrt3/2)=((5+sqrt3)/2, (1+sqrt3)/2)
$\implies P R = \frac{5 + \sqrt{3}}{2} - 0 = \frac{5 + \sqrt{3}}{2}$
$\implies P Q = \frac{P R}{\cos} 45 = \frac{5 + \sqrt{3}}{2} \cdot \frac{2}{\sqrt{2}} = \frac{5 + \sqrt{3}}{\sqrt{2}}$
$\implies P {Q}^{2} = {\left(\frac{5 + \sqrt{3}}{\sqrt{2}}\right)}^{2} = 14 + 5 \sqrt{3}$

Apr 9, 2018

Given the equation of the circle ${x}^{2} + {y}^{2} - 5 x - y + 5 = 0$.

The standard form of this equation :

${x}^{2} - 2 \cdot x \cdot \frac{5}{2} + {\left(\frac{5}{2}\right)}^{2} + {y}^{2} - 2 \cdot x \cdot \frac{1}{2} + {\left(\frac{1}{2}\right)}^{2} = \frac{25}{4} + \frac{1}{4} - 5 = \frac{3}{2}$

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

So coordinates of its center is $\left(\frac{5}{2} , \frac{1}{2}\right)$ and its radius $r = \sqrt{\frac{3}{2}}$

So coordinates of any point $\left(Q\right)$ on the circle may be written parametrically as $\left(\frac{5}{2} + r \cos \theta , \frac{1}{2} + r \sin \theta\right)$

The coordinates of given point $\left(P\right)$ is $\left(0 , - 2\right)$.

So

$P {Q}^{2} = {\left(\frac{5}{2} + r \cos \theta\right)}^{2} + {\left(\frac{1}{2} + r \sin \theta + 2\right)}^{2}$

$\implies P {Q}^{2} = {\left(\frac{5}{2} + r \cos \theta\right)}^{2} + {\left(\frac{5}{2} + r \sin \theta\right)}^{2}$

$\implies P {Q}^{2} = \frac{25}{4} + {r}^{2} {\cos}^{2} \theta + \frac{25}{4} + {r}^{2} {\sin}^{2} \theta + 2 \cdot \frac{5}{2} \cdot r \cos \theta + 2 \cdot \frac{5}{2} \cdot \sin \theta$

$\implies P {Q}^{2} = \frac{25}{2} + {r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + 5 r \left(\cos \theta + \sin \theta\right)$

$\implies P {Q}^{2} = \frac{25}{2} + {\left(\sqrt{\frac{3}{2}}\right)}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + 5 \sqrt{\frac{3}{2}} \cdot \sqrt{2} \left(\frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta\right)$

$\implies P {Q}^{2} = \frac{25}{2} + \frac{3}{2} \cdot 1 + 5 \sqrt{\frac{3}{2}} \cdot \sqrt{2} \left(\sin \left(\frac{\pi}{4}\right) \cos \theta + \cos \left(\frac{\pi}{4}\right) \sin \theta\right)$

$\implies P {Q}^{2} = 14 + 5 \sqrt{3} \sin \left(\theta + \frac{\pi}{4}\right)$

$\implies {\left(P {Q}^{2}\right)}_{\text{max}} = 14 + 5 \sqrt{3}$, as maximum value of $\sin \left(\theta + \frac{\pi}{4}\right)$ is $1$