(In reply to
Solution by Brian Smith)
My solution was very similar, I just thought of things a bit differently.
Simplify the second equation as you did but then see that
√(x2+y2-4x+2y+5)=a and
sqrt[(x-10)^2 + (y-5)^2]=b
Where a+b=10 would be two circles (a>0, b>0 of course)
one at (2,-1) with radius a and the other at (10,5).
These centers are also 10 apart, so they always have a common tangent on the line segment connecting their centers.
The rest is the same.
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Posted by Jer
on 2023-07-16 15:23:35 |