$\text{ If }x,y\in R\text{ and }15x+8y=34,\text{ then }{\left(x^2+y^2\right)}_{min}\text{ , is equal to}$

$15x+8y=34----\left(1\right)$

$\text{ Let }P(x,y)\text{ be any point on line (i). }$

$\text{ Now, }\sqrt{x^2+y^2}=\text{ Distance of point }P(x,y)\text{ from origin }$

$\text{ Also, perpendicular distance of line (i) from }(0,0)$

$=\frac{|-34|}{\sqrt{{15}^2+8^2}}=2\text{ units }$

$\text{ This is the shortest distance of }P\text{ , lying on (i) and }(0,0)\text{ . }$

$\begin{array}{r}\sqrt{x^2+y^2}\ge 2\\ \Rightarrow x^2+y^2\ge 4\end{array}$

$\text{ Therefore, minimum value of }\left(x^2+y^2\right)\text{ is }4\text{ . }$