document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=202522>Question 202522</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Describe in words the graph of each of these curves below. Include in your description the shape, along with other possible relevant information such as length, width, and center points.\r<BR>\n" );
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document.write( "a. (X+2)2 + (Y-4)2 = 36\r<BR>\n" );
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document.write( "I need help learning to solve for x and y (those are squared by the way) and how to find the slope so I can graph it. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #152735 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=152735\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Well, in this type of problem (identifying curves from their equations), it is helpful to recognise the standard form of the equation for a circle with its centre at (h, k) and having a radius of r:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2 ALIGN=MIDDLE  ALT=\"%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "Do you see the similarity between this and the given equation?<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%5E2%2B%28y-2%29%5E2+=+36 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%5E2%2B%28y-2%29%5E2+=+36\"   DISABLED_style= \"max-width:90%; height:auto;\" >: h = -2, k = 2, and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%5E2+=+36 ALIGN=MIDDLE  ALT=\"r%5E2+=+36\"   DISABLED_style= \"max-width:90%; height:auto;\" > so that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+6 ALIGN=MIDDLE  ALT=\"r+=+6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Can you take it from here?<BR>\n" );
document.write( "By the way, you do not have to solve for x or y to answer the question.<BR>\n" );
document.write( "But you will have to solve for y to graph the equation.<BR>\n" );
document.write( "Here's how you would do this:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%5E2%2B%28y-2%29%5E2+=+36 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%5E2%2B%28y-2%29%5E2+=+36\"   DISABLED_style= \"max-width:90%; height:auto;\" > Subtract <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%5E2 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > from both sides of the equation.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28y-2%29%5E2+=+-%28x%2B2%29%5E2%2B36 ALIGN=MIDDLE  ALT=\"%28y-2%29%5E2+=+-%28x%2B2%29%5E2%2B36\"   DISABLED_style= \"max-width:90%; height:auto;\" > Now take the square root of both sides.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-2+=+sqrt%28-%28x%2B2%29%5E2%2B36%29 ALIGN=MIDDLE  ALT=\"y-2+=+sqrt%28-%28x%2B2%29%5E2%2B36%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Add 2 to both sides to isolate the y.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+2%2B-sqrt%28-%28x%2B2%29%5E2%2B36%29 ALIGN=MIDDLE  ALT=\"y+=+2%2B-sqrt%28-%28x%2B2%29%5E2%2B36%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > You can simplify the radicand (the quantity under the radical sign) a bit.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+2%2B-sqrt%28-x%5E2-4x%2B32%29 ALIGN=MIDDLE  ALT=\"y+=+2%2B-sqrt%28-x%5E2-4x%2B32%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Notice that you really have two answers here:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+2%2Bsqrt%28-x%5E2-4x%2B32%29 ALIGN=MIDDLE  ALT=\"y+=+2%2Bsqrt%28-x%5E2-4x%2B32%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+2-sqrt%28-x%5E2-4x%2B32%29 ALIGN=MIDDLE  ALT=\"y+=+2-sqrt%28-x%5E2-4x%2B32%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > and, to graph this, you would have to graph both parts, one for the upper half of the circle and the other for the lower half.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28400%2C400%2C-10%2C5%2C-5%2C10%2C2%2Bsqrt%28-x%5E2-4x%2B32%29%2C2-sqrt%28-x%5E2-4x%2B32%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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