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Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
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 $Q$ .
$\text{max}\left(\sqrt{x^{2} + y^{2}}\right)=OP+PQ$
$=\sqrt{\left(8 - 0\right)^{2} + \left(- 15 - 0\right)^{2}}+3$
$=17+3=20$
