Question 258000
Second equation is a circle, so reorganise it
(x^2 + 12 x) - (y^2 - 6y)+ 29 = 0 and then complete the square
(x^2 + 12x + 36) - ( y^2 - 6y + 9) - 45 + 29 = 0 which becomes
(x + 6)^2 - (y - 3)^2 = 4^2 which is a circle centre at -6, 3 and radius 4
other equation is y = (1/2)x - 4 or x = 2y + 8
Probably solve for y is easier
(2y + 14)^2 - (y - 3)^2 = 4^2
4y^2 + 56 y + 196 -(y^2 - 6y +9) = 16
3y^2 + 62y + 187 = 16
3y^2 + 62y + 171 = 0*[invoke quadratic "y", 3, 62, 171 ]
Then use x = 2y + 8 to get the x figures