Lesson Transform general equation of a circle to the standard form by completing the squares

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<H2>Transform general equation of a circle to the standard form by completing the squares</H2>

This lesson shows you by examples on how to transform a general equation of a circle to the standard form by completing the squares.

<H3>Problem 1</H3>Transform a general equation of a circle &nbsp;{{{x^2 + y^2 - 2x + 6y - 12}}} = {{{0}}} &nbsp;to the standard form. 
Then identify its center and the radius.

<B>Solution</B>

<pre>
{{{x^2 + y^2 - 2x + 6y - 12}}} = {{{0}}}   --->   (move the constant term to the right, if necessary)  --->

{{{x^2 + y^2 - 2x + 6y}}} = {{{12}}}     --->   (group the terms containing x and y separately)  --->

{{{(x^2-2x) + (y^2+6y)}}} = {{{12}}}  --->   (Add constant terms to x-group and y-group to complete the squares)

{{{(x^2-2*x + 4) + (y^2+2*3y+9)}}} = {{{12+4+9}}},   (add the same constant terms to the right side to keep the balance)  --->
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 <TR>
 <TD>
{{{(x-2)^2 + (y+3)^2}}} = {{{25}}},

{{{(x-2)^2+ (y+3)^2}}} = {{{5^2}}}.

Thus you got the equation of the circle in the standard form. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

The center of the circle is the point  (2,-3).

The radius of the circle is r = 5 units.

</TD>
  <TD>
{{{drawing( 300, 300, -4.5, 8.5, -9.5, 3.5,
            grid(1),
            circle(2, -3, 5),
            circle(2, -3, 0.2)
)}}}

Circle {{{(x-2)^2}}} + {{{(y+3)^2}}} = 25
  </TD>
  </TR>
</TABLE></pre><H3>Problem 2</H3>Transform a general equation of a circle &nbsp;{{{2x^2 + 2y^2 + 4x + 8y - 14}}} = {{{0}}} &nbsp;to the standard form. 
Then identify its center and the radius.

<B>Solution</B>

<pre>
{{{2x^2 + 2y^2 + 4x + 8y - 14}}} = {{{0}}}  --->  (Divide both sides by 2 to get the leading coefficient at {{{x^2}}} and {{{y^2}}} equal to 1)  --->

{{{x^2 + y^2 + 2x + 4y - 7}}} = {{{0}}}   --->   (move the constant term to the right, if necessary)  --->

{{{x^2 + y^2 + 2x + 4y}}} = {{{7}}}     --->   (group the terms containing x and y separately)  --->

{{{(x^2+2x) + (y^2+4y)}}} = {{{7}}}  --->   (Add constant terms to x-group and y-group to complete the squares)

{{{(x^2+2*x + 1) + (y^2+2*2y+4)}}} = {{{7+1+4}}},   (add the same constant terms to the right side to keep the balance)  --->
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 <TR>
 <TD>
{{{(x+1)^2 + (y+2)^2}}} = {{{12}}},

{{{(x+1)^2+ (y+2)^2}}} = {{{(sqrt(12))^2}}}.

Thus you got the equation of the circle in the standard form. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

The center of the circle is the point  (1,2).

The radius of the circle is r = {{{sqrt(12)}}} = {{{2*sqrt(3)}}} units.

</TD>
  <TD>
{{{drawing( 300, 300, -3.5, 5.5, -2.5, 6.5,
            grid(1),
            circle(1, 2, sqrt(12)),
            circle(1, 2, 0.15)
)}}}

Circle {{{(x-1)^2}}} + {{{(y-2)^2}}} = 12
  </TD>
  </TR>
</TABLE></pre>

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