(-7,6) is a point in Quadrant II
and if theta is the angle between the positive X-axis and (-7,6) with vertex at the origin:
color(white)("XXX")tan(theta)=-6/7
Using a claculator
color(white)("XXX")hattheta=arctan(-6/7)
gives us
color(white)("XXX")hattheta = -0.70863
hattheta is the same reference angle as the required theta but since arctan(_) is restricted to a range of (-pi/2,pi/2]
hattheta is in Quadrant IV
Therefore
color(white)("XXX")theta = pi+arctan(-6/7) ~= 2.43
r (the radius) can be calculated using the Pythagorean Theorem as
color(white)("XXX")r=sqrt((-7)^2+6^2) ~= 9.22