Question

Let$A(1,4)$ and$B(1,-5)$be two points. Let $P$ be a point on the circle${(x-1)}^2+{(y-1)}^2=1$such that ${\left(PA\right)}^2+{\left(PB\right)}^2$ have maximum value, then the points $P,A\text{and}B$ lie on.

• A. a parabola
• B. a straight line
• C. a hyperbola
• D. an ellipse

Answer 1

The correct option is B

a straight line

Explanation for the correct option:

Step 1: Finding the value of ${\left(PA\right)}^2+{\left(PB\right)}^2$

We know that points on the circle ${(x-a)}^2+{(y-b)}^2=r^2$ are in the form of $\left(r\cos \theta +a),(r\sin \theta +b\right)$

Given, $P$ be a point on the circle${(x-1)}^2+{(y-1)}^2=1$

Therefore, $P$= $=\left(1\cos \theta +1,1\sin \theta +1\right)$

Finding ${\left(PA\right)}^2+{\left(PB\right)}^2$

Using distance formula $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

$P\left(\left(\cos \theta +1\right),\left(\sin \theta +1\right)\right)$,$A(1,4)$,$B(1,-5)$

$$\begin{gathered} {\left(PA\right)}^2={\left(\left(\cos \theta +1\right)-1\right)}^2+{\left(\left(\sin \theta +1\right)-4\right)}^2 \\ ={\left(\cos \theta \right)}^2+{\left(\sin \theta -3\right)}^2 \\ ={\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)-6\sin \theta +9 \\ =1-6\sin \theta +9\left[\because co{s}^2\left(\theta \right)+si{n}^2\left(\theta \right)=1\right] \\ =10-6\sin \theta \end{gathered}$$

$$\begin{gathered} {\left(PB\right)}^2={\left(\left(\cos \theta +1\right)-1\right)}^2+{\left(\left(\sin \theta +1\right)+5\right)}^2 \\ ={\left(\cos \theta \right)}^2+{\left(\sin \theta +6\right)}^2 \\ ={\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)+12\sin \theta +36 \\ =1+12\sin \theta +36\mathbf{}\left[\because co{s}^2\left(\theta \right)+si{n}^2\left(\theta \right)=1\right] \\ =37+12\sin \theta \end{gathered}$$

$$\begin{gathered} {\overline{){\left(PA\right)}^2+{\left(PB\right)}^2}}_{max}=10-6\sin \theta +37+12\sin \theta \\ =47+6{\overline{)\sin \theta }}_{max} \end{gathered}$$

when $\theta =\frac{\mathrm{\pi }}{2},$$\sin$ attains maximum value,

Step 2: Finding the value of $x$ on which the points lie:

Substitute $\theta =\frac{\mathrm{\pi }}{2}$in $P$

$\begin{array}{rcl}\left(\left(\cos \frac{\pi }{2}+1\right),\left(\sin \frac{\pi }{2}+1\right)\right)& =& \left(0+1,1+1\right)\\ & =& \left(1,2\right)\end{array}$

Therefore $P\left(1,2\right),A\left(1,4\right)\text{and}B\left(1,-5\right)$ are collinear and lie on $x=1$

Hence, option (B) is the correct answer.