Question 264546
If you accept that they intersect at A and B, then all you need to show is that the distance from A to B is the diameter.
{{{r^2 = 20}}} so {{{d^2 = 80}}}
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The square of the distance from A to B is
{{{diffy^2 + diffx^2}}}
= {{{4^2 + 8^2}}}
= 80
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It's the same length as the diameter, so it's thru the center of the circle.
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It was stated that A and B are on the circle and the line, so no proof of that was done.