Question 153717


What is the graph of the equation x2 + y2 - 8x + 6y + 24 = 0

This is a circle because {{{x^2}}} and {{{y^2}}} have the
same coefficient when on the same side of the equation.


{{{x^2 + y^2 - 8x + 6y + 24 = 0}}}
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We must first get the equation in standard form:

{{{(x-h)^2+(y-k)^2= r^2}}} 

Rearrange to get the x terms together and the y terms togethsr.
That is, swap the middle two terms:

{{{x^2 - 8x + y^2 + 6y + 24 = 0}}}


{{{(x^2 - 8x) + (y^2 + 6y) + 24 = 0}}}

Out to the side or on scratch paper,

we complete the square of

{{{(x^2-8x)}}}:

1. Multiply the coefficient, -8, of {{{x}}}, by {{{1/2}}},
   getting {{{-4}}}.
2. Then square {{{-4}}}, getting +16
3. Add +16 to the left side inside the first parentheses, and
   also add it to the right side:

{{{(x^2-8x+16) + (y^2 + 6y) + 24= 0+16}}}   

Now we complete the square of

{{{(y^2+6y)}}}:

1. Multiply the coefficient, 6, of {{{y}}}, by {{{1/2}}},
   getting {{{3}}}.
2. Then square {{{3}}}, getting +9
3. Add +9 to the left side inside the first parentheses, and
   also add it to the right side:

{{{(x^2 - 8x+16) + (y^2 + 6y+9) + 24= 0+16+9}}} 

Factor each parentheses:

{{{(x-4)(x-4) + (y+3)(y+3) +24=25}}}

Write as perfect squares:

{{{(x-4)^2 + (y+3)^2 +24 = 25}}}

Add -24 to both sides:

{{{(x-4)^2 + (y+3)^2 = 1}}}

Compare to the standard form:

{{{(x-h)^2 + (y-k)^2 = r^2}}}

This has center (h,k) and radius r.

{{{-h=-4}}} so {{{h=4}}}, {{{-k=+3}}}, so {{{k=-3}}}.  {{{r^2=1}}}, so {{{r=1}}}

so the center, (h,k) = (4,-3) and the radius is 1.

So we plot the center (4,-3).

{{{drawing(300,300,-2,6,-6,2, locate(4-.1,-3+.4-.2,"@"),
 graph(300,300,-2,6,-6,2)
 
)}}}

Get a compass and open it one unit wide.  Place the sharp 
point of the compass on the center and draw the circle.
That is the graph:

{{{drawing(300,300,-2,6,-6,2, locate(4-.1,-3+.4-.2,"@"),
circle(4,-3,1), graph(300,300,-2,6,-6,2)
 
)}}}

Edwin</pre>