The algorithm is tripping! It won't let me post a legit explanation. Sorry!
Find the equation of the circle that has a diameter with endpoint located at (-5,-3) and (-11,-3).
1 answer:
Check the picture below.
you can pretty much, without having to use the distance formula, see the diameter is 6 units long, thus the radius is half that or 3 then.
now, without having to use the midpoint formula, if we move from either endpoint inwards 3 units, we'll end up at the center, -5 -3, is just -8, so the center will be at -8, -3.
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad center~~(\stackrel{-8}{{{ h}}},\stackrel{-3}{{{ k}}})\qquad \qquad radius=\stackrel{3}{{{ r}}} \\\\\\\ [x-(-8)]^2+[y-(-3)]^2=3^2\implies (x+8)^2+(y+3)^2=9](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-8%7D%7B%7B%7B%20h%7D%7D%7D%2C%5Cstackrel%7B-3%7D%7B%7B%7B%20k%7D%7D%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B3%7D%7B%7B%7B%20r%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-8%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D3%5E2%5Cimplies%20%28x%2B8%29%5E2%2B%28y%2B3%29%5E2%3D9)
you can pretty much, without having to use the distance formula, see the diameter is 6 units long, thus the radius is half that or 3 then.
now, without having to use the midpoint formula, if we move from either endpoint inwards 3 units, we'll end up at the center, -5 -3, is just -8, so the center will be at -8, -3.
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