Lösung 4.1:5b
Aus Online Mathematik Brückenkurs 1
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Version vom 14:46, 22. Okt. 2008
If the circle is to contain the point (-1,1), then that point's distance away from the centre (2,-1) must equal the circle's radius, r. Thus, we can obtain the circle's radius by calculating the distance between (-1,1) and (2,-1) using the distance formula,
| \displaystyle \begin{align}
r &= \sqrt{(2-(-1))^2+(-1-1)^2} = \sqrt{3^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\,\textrm{.} \end{align} |
When we know the circle's centre and its radius, we can write the equation of the circle,
| \displaystyle (x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2} |
which the same as
| \displaystyle (x-2)^{2} + (y+1)^2 = 13\,\textrm{.} |
Note: A circle having its centre at (a,b) and radius r has the equation
| \displaystyle (x-a)^2 + (y-b)^2 = r^2\,\textrm{.} |
