Question

lf a line passes through the point $P(1,-2)$ and cuts the circle $x^2+y^2-x-y=0$ at $A$ and $B$, then the maximum value of $PA+PB$ is

• A. $\sqrt{26}$
• B. $8$
• C. $\sqrt{8}$
• D. $2\sqrt{8}$

Answer 1

Answer: (A)

$\sqrt{26}$

Given equation of the circle: $x^2+y^2-x-y=0$

For maximum value of $PA+PB$, the line $PAB$ must pass through centre of the circle.

$AC=OC=BC=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}$

$=\frac{1}{\sqrt{2}}$

and $PC=\sqrt{{(1-\frac{1}{2})}^2+{(-2-\frac{1}{2})}^2}$

$=\frac{\sqrt{26}}{2}$

Now, $PA=PC-AC=\frac{\sqrt{26}}{2}-\frac{1}{\sqrt{2}}$

and $PB=PC+BC=\frac{\sqrt{26}}{2}+\frac{1}{\sqrt{2}}$

Now, $PA+PB=\frac{\sqrt{26}}{2}-\frac{1}{\sqrt{2}}+\frac{\sqrt{26}}{2}+\frac{1}{\sqrt{2}}=\sqrt{26}$

Hence, option A is the correct option.