Solution 4.1:7b
From Förberedande kurs i matematik 1
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| - | {{ | + | The equation is almost in the standard form for a circle; all that is needed is for us to collect together the ''y''²- and ''y''-terms into a quadratic term by completing the square |
| - | <center> [[ | + | |
| - | + | {{Displayed math||<math>y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.}</math>}} | |
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| + | After rewriting, the equation is | ||
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| + | {{Displayed math||<math>x^2 + (y+2)^2 = 4</math>}} | ||
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| + | and we see that the equation describes a circle having its centre at (0,-2) and radius <math>\sqrt{4}=2\,</math>. | ||
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| + | <center> [[Image:4_1_7_b.gif]] </center> | ||
Current revision
The equation is almost in the standard form for a circle; all that is needed is for us to collect together the y²- and y-terms into a quadratic term by completing the square
| \displaystyle y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.} |
After rewriting, the equation is
| \displaystyle x^2 + (y+2)^2 = 4 |
and we see that the equation describes a circle having its centre at (0,-2) and radius \displaystyle \sqrt{4}=2\,.
