Question

If a point P has coordinates $\left( {0, - 2} \right)$ and Q is any point on the circle ${x^2} + {y^2} - 5x - y + 5 = 0$ , then the maximum value of ${\left( {PQ} \right)^2}$ is
A) $8 + 5\sqrt 3 $
B) $\dfrac{{47 + 10\sqrt 6 }}{2}$
C) $14 + 5\sqrt 3 $
D) $\dfrac{{25 + \sqrt 6 }}{2}$

Answer 1

Hint:
Firstly, find a, b and r by simplifying the equation ${x^2} + {y^2} - 5x - y + 5 = 0$ and bringing it to the form $\left( {{x^2} - {a^2}} \right) + \left( {{y^2} - {b^2}} \right) = {r^2}$ .
Then, the coordinates of point Q on the circle are given by $\left( {a + r\cos Q,b + r\sin Q} \right)$.
Finally, for the value of ${\left( {PQ} \right)^2}$ , we will find the distance between the points P and Q using the distance formula ${\left( {PQ} \right)^2} = {\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2}$.

Complete step by step solution:
Here, it is given that Q is any point on the circle ${x^2} + {y^2} - 5x - y + 5 = 0$ .
Now, we will be simplifying the equation of circle by making perfect square polynomials as follows
 \[{x^2} - 5x + \dfrac{{25}}{4} - \dfrac{{25}}{4} + {y^2} - y + \dfrac{1}{4} - \dfrac{1}{4} + 5 = 0\]
 \[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} - \left( {\dfrac{{25 + 1}}{4}} \right) + 5 = 0\]
 \[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} - \dfrac{{26}}{4} + 5 = 0\]
 \[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} - \dfrac{{13}}{2} + 5 = 0\]
 \[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} - \left( {\dfrac{{13 - 10}}{2}} \right) = 0\]
 \[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} = \dfrac{3}{2}\]
 \[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} + {\left( {y - \dfrac{1}{2}} \right)^2} = {\left( {\sqrt {\dfrac{3}{2}} } \right)^2}\]
Thus, by comparing the above equation with $\left( {{x^2} - {a^2}} \right) + \left( {{y^2} - {b^2}} \right) = {r^2}$ , we get center of the circle as $\left( {a,b} \right) = \left( {\dfrac{5}{2},\dfrac{1}{2}} \right)$ and radius $r = \dfrac{{\sqrt 3 }}{2}$ .
Now, coordinates of the point Q can be written as $\left( {a + r\cos Q,b + r\sin Q} \right) = \left( {\dfrac{5}{2} + \dfrac{{\sqrt 3 }}{2}\cos Q,\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}\sin Q} \right)$ .
Also, it is given that a point P has coordinates $\left( {0, - 2} \right)$ .
          

Now, for ${\left( {PQ} \right)^2}$ , we have to find the distance between the points P and Q, which can be given by
 $
  {\left( {PQ} \right)^2} = {\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2} \\
   = {\left( {0 - \dfrac{5}{2} - \sqrt {\dfrac{3}{2}} \cos Q} \right)^2} + {\left( { - 2 - \dfrac{1}{2} - \sqrt {\dfrac{3}{2}} \sin Q} \right)^2} \\
   = \dfrac{{25}}{4} + 2\left( {\dfrac{5}{2}} \right)\left( {\sqrt {\dfrac{3}{2}} \cos Q} \right) + \dfrac{3}{4}{\cos ^2}Q + \dfrac{{25}}{4} + 2\left( {\dfrac{5}{2}} \right)\left( {\sqrt {\dfrac{3}{2}} \sin Q} \right) + \dfrac{3}{4}{\sin ^2}Q \\
   = \dfrac{{25}}{2} + \dfrac{3}{2} + 5\sqrt {\dfrac{3}{2}} \cos Q + 5\sqrt {\dfrac{3}{2}} \sin Q \\
   = \dfrac{{25}}{2} + \dfrac{3}{2} + 5\sqrt {\dfrac{3}{2}} \left( {\cos Q + \sin Q} \right) \\
 $
Now, for the maximum value of ${\left( {PQ} \right)^2}$ , the values of cosine and sine functions must be equal. So, $\cos Q = \sin Q$ .
 $
  \cos Q = \sin Q \\
   \Rightarrow 1 = \dfrac{{\sin Q}}{{\cos Q}} \\
   \Rightarrow \tan Q = 1 \\
   \Rightarrow \tan Q = \tan 45^\circ \\
   \Rightarrow \angle Q = 45^\circ \\
 $
 $ \Rightarrow \cos 45^\circ = \sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$
 \[
   \Rightarrow {\left( {PQ} \right)^2} = \dfrac{{25}}{2} + \dfrac{3}{2} + 5\sqrt {\dfrac{3}{2}} \left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}} \right) \\
   = \dfrac{{25}}{2} + \dfrac{3}{2} + 5\sqrt {\dfrac{3}{2}} \left( {\dfrac{2}{{\sqrt 2 }}} \right) \\
   = \dfrac{{28}}{2} + 5\dfrac{{\sqrt 3 }}{{\sqrt 2 }}\left( {\sqrt 2 } \right) \\
   = 14 + 5\sqrt 3 \\
 \]
So, option (C) is correct.

Note:
The above given question can be solved by the help of the general equation of a circle with a center other than origin and by the help of the distance formula for a two-dimensional line. By drawing the correct figure and using the distance formula, we may get the correct answer.
The general equation of a circle having center at any point is $\left( {{x^2} - {a^2}} \right) + \left( {{y^2} - {b^2}} \right) = {r^2}$ and the distance formula between any two points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ is ${\left( {AB} \right)^2} = {\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2}$.