Question 414573
x^2+y^2=808
x=y+4
substitute the value of x in the equation
(y+4)^2+y^2=808
y^2+8y+16+y^2=808
2y^2+8y-792=0
/2
y^2+4y-396=0
y^2+22y-18y-396=0
y(y+22)-18(y+22)=0
(y+22)(y-18)=0
y=-22 OR 18

x=-18 OR 22

(-18,-22), (18,22)