At what point in the third quadrant does the line with equation y = x + 2 intersect the circle with radius 6 and center (0,2)?

Answer 1

The key to this problem is noting that the center is on the line.

Then, as the slope is 1, and the radius is 6, the line intersects the circle at the center + (3√2,3√2) and the center - (3√2,3√2) The third quadrant value will then be (0, 2) - (3√2,3√2) = (-3√2,2-3√2)