OpenSAT

The equation of the circle with center at the origin and radius 5 is $x^2 + y^2 = 25$. We can solve the system of equations $x^2 + y^2 = 25$ and $x + y = 7$ to find the points of intersection. Solving for $x$ in the second equation, we get $x = 7 - y$. Substituting this into the first equation, we get $(7-y)^2 + y^2 = 25$. Expanding the equation, we get $49 - 14y + y^2 + y^2 = 25$. Combining like terms, we get $2y^2 - 14y + 24 = 0$. Dividing both sides by 2, we get $y^2 - 7y + 12 = 0$. Factoring, we get $(y-3)(y-4) = 0$. Therefore, $y = 3$ or $y = 4$. Substituting these values back into the equation $x = 7 - y$, we get the points of intersection as $(4,3)$ and $(3,4)$. The distance between these two points can be found using the distance formula: $\sqrt{(4-3)^2 + (3-4)^2} = \sqrt{1+1} = \sqrt{2}$. Thus, the distance between the two points is $5\sqrt{2}$.