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Rearrange a homogeneous circle equation into the form (x-xc)^2+(y-yc)^2=r^2 to find the centre and radius.

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If

\n

$\\var{eqn}$

\n

obtain the centre and radius of the circle that this equation represents.

", "advice": "

The original equation $\\var{eqn}$ becomes

\n

$(\\var{simplify(expression(\"x^2-2*x*\"+Cx),\"all\")}+\\var{Cx^2})-\\var{Cx^2} + (\\var{simplify(expression(\"y^2-2*y*\"+Cy),\"all\")}+\\var{Cy^2})-\\var{Cy^2}$  $+$ $\\var{const}$ $=0$

\n

$\\var{simplify(expression(\"(x-\"+Cx+\")^2\"),\"all\")}+\\var{simplify(expression(\"(y-\"+Cy+\")^2\"),\"all\")}-\\var{Cx^2}-\\var{Cy^2}$ $+$ $\\var{const}$ $=0$

\n

$\\therefore\\quad \\var{simplify(expression(\"(x-\"+Cx+\")^2\"),\"all\")}+\\var{simplify(expression(\"(y-\"+Cy+\")^2\"),\"all\")}=\\var{R2}$

\n

which represents a circle with centre $(\\var{Cx},\\var{Cy})$ and radius $\\var{R}$.

\n

{graph}

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Centre x value

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The centre y-value

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The square of the radius

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Begin by completing the square separately on the $x-$terms and the $y-$terms.

\n

$\\var{simplify(expression(\"x^2-2*\"+Cx+\"*x\"),\"all\")} = (x + $[[0]]$)^2$ - [[1]]

\n

$\\var{simplify(expression(\"y^2-2*\"+Cy+\"*y\"),\"all\")} = (y + $[[2]]$)^2$ - [[3]]

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Now complete the problem:

\n

centre = ([[0]],[[1]])

\n

radius = [[2]]

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