Question

If real numbers $x$ and $y$ satisfy $x^2+y^2-16x+30y+280=0,$ then maximum value of ${\left(x^2+y^2\right)}^{1/2}$ is

• A. $15$
• B. $20$
• C. $25$
• D. $30$

Answer 1

Answer: (B)

We have,
$x^2+y^2-16x+30y+280=0$
$\Rightarrow {\left(x-8\right)}^2+{\left(y+15\right)}^2={\left(3\right)}^2$
Centre $\equiv \left(8,-15\right)$ , Radius $=3$
Now, $\sqrt{x^2+y^2}$ represents the distance of $\left(x,y\right)$ from the origin.

Distance will be maximum along the diameter at the point $Q$ .
$\text{max}\left(\sqrt{x^2+y^2}\right)=OP+PQ$
$=\sqrt{{\left(8-0\right)}^2+{\left(-15-0\right)}^2}+3$
$=17+3=20$