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      <text>17.1.1.5 Equation of a circle</text>
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<h4 class="HELM_example">Example</h4>
<p>Show that the expression \[ x^2+y^2-2x+6y+6=0\] represents the equation of a circle. Find its centre and radius.</p>
<h4 class="HELM_solution">Solution</h4>
<p>We shall see later how to recognise this as the equation of a circle simply by examination of the coefficients of the quadratic terms \(x^2,\ y^2\) and \(xy\). However, in the present example we will use the process of completing the square, for \(x\) and for \(y\), to show that the expression can be written in standard form. 
  
Now \(x^2+y^2-2x+6y+6 = x^2-2x+y^2+6y+6\). 
Also,
  \[x^2-2x = (x-1)^2-1\ \ \text{and}\ \ y^2+6y = (y+3)^2-9.\]
Hence we can write 
  \[x^2+y^2-2x+6y+6 = (x-1)^2-1+(y+3)^2-9+6=0\]
or, taking the free constants to the right-hand side: 
  \[(x-1)^2+(y+3)^2=4.\]
  By comparing this with the standard form we conclude this represents the equation of a circle with centre at \((1,-3)\) and radius \(2\).</p>]]></text>
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