they haven't used pythagoras at all, as we know that the line $$ l_1$$ is parallel to the x axis the gradient of $$l_1$$ is going to be 0 so essentially just a straight line through the circle, using the fact that the distance of PQ is 8 we know that from the x co ordinate of the centre the x co ordinates points P and Q will be $$9 \pm \frac{8}{2}$$ as it will be half the distance from PQ now you can plug the x values into the equation of the circle and find that we get $$(-4)^2 + (y+6)^2 = 36$$ and $$ (4)^2 + (y+6)^2 = 36$$ as these are the same quadratic we solve one and find that $$y = -6\pm 2 \sqrt{5}$$
#part b
outer stump
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KingDavid
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they haven't used pythagoras at all, as we know that the line $$ l_1$$ is paral...
yhh i get it now ty