The intersection of straight #s_1->y+x=6# and it's normal #s_2# passing by #p_c={x_c,y_c} = {0,0}# the circumference center, is also the tangency point between the circumference and the straight #s_1#.

The #s_2# equation can be obtained knowing that #m_1# is the #s_1# declivity. Then

#s_2->y = y_c -(1/m_1)(x-x_c)#

but #m_1 = ((dy)/(dx))_{s_1} = -1#

so

#s_2->y = 0 -(1/(-1))(x-0) = x#

#s_1 nn s_2->{(y +x = 6),(y - x = 0):}->{x_t=3,y_t=3}#

where #p_t = {x_t,y_t}# is the tangency point.

we know that #r = norm(p_c-p_t) = 3sqrt(2)#

The circumference equation reads

#(x-x_c)^2+ (y-y_c)^2=r^2#

or

#x^2+y^2=18#