Solution 4.1:7a

From Förberedande kurs i matematik 1

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Current revision (11:19, 8 October 2008) (edit) (undo)
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As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine
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As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine ''x''- and ''y''-terms together in their own respective square terms, then we will have the equation in the standard form
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<math>x</math>
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<math>y</math>
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- terms together in their own respective square terms, then we will have the equation in the standard form,
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<math>\left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}</math>
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{{Displayed math||<math>(x-a)^2 + (y-b)^2 = r^2\,,</math>}}
and we will then be able to read off the circle's centre and radius.
and we will then be able to read off the circle's centre and radius.
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If we take the
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If we take the ''x''- and ''y''-terms on the left-hand side and complete the square, we get
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<math>x</math>
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<math>y</math>
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- terms on the left-hand side and complete the square, we get
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{{Displayed math||<math>\begin{align}
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<math>x^{2}+2x=\left( x+1 \right)^{2}-1^{2}</math>
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x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt]
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y^2 - 2y &= (y-1)^2-1^2\,,
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\end{align}</math>}}
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<math>y^{2}-2y=\left( y-1 \right)^{2}-1^{2}</math>
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and then the whole equation can be written as
and then the whole equation can be written as
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{{Displayed math||<math>(x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,</math>}}
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<math>\left( x+1 \right)^{2}-1^{2}+\left( y-1 \right)^{2}-1^{2}=1</math>
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or, with the constants moved to the right-hand side,
or, with the constants moved to the right-hand side,
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{{Displayed math||<math>(x+1)^2 + (y-1)^2 = 3\,\textrm{.}</math>}}
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<math>\left( x+1 \right)^{2}+\left( y-1 \right)^{2}=3</math>
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This is a circle having its centre at (-1,1) and radius <math>\sqrt{3}\,</math>.
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This is a circle having its centre at
 
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<math>\left( -1 \right.,\left. 1 \right)</math>
 
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and radius
 
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<math>\sqrt{3}</math>.
 
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{{NAVCONTENT_START}}
 
<center> [[Image:4_1_7a-2(2).gif]] </center>
<center> [[Image:4_1_7a-2(2).gif]] </center>
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{{NAVCONTENT_STOP}}
 

Current revision

As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine x- and y-terms together in their own respective square terms, then we will have the equation in the standard form

\displaystyle (x-a)^2 + (y-b)^2 = r^2\,,

and we will then be able to read off the circle's centre and radius.

If we take the x- and y-terms on the left-hand side and complete the square, we get

\displaystyle \begin{align}

x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt] y^2 - 2y &= (y-1)^2-1^2\,, \end{align}

and then the whole equation can be written as

\displaystyle (x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,,

or, with the constants moved to the right-hand side,

\displaystyle (x+1)^2 + (y-1)^2 = 3\,\textrm{.}

This is a circle having its centre at (-1,1) and radius \displaystyle \sqrt{3}\,.


Image:4_1_7a-2(2).gif
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