If x,y∈R and 15x+8y=34, then x2+y2min , is equal to

15x+8y=34----(1) Let P(x,y) be any point on line (i).  Now, x2+y2= Distance of point P(x,y) from origin  Also, perpendicular distance of line (i) from (0,0)=|−34|152+82=2 units  This is the shortest distance of P , lying on (i) and (0,0) . x2+y2≥2⇒ x2+y2≥4 Therefore, minimum value of x2+y2 is 4 .