document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=756440>Question 756440</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The equation of the circle with center at (-2,3) and tangent to the y-axis is\r<BR>\n" );
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document.write( "	(x+2)^2+(y-3)^2=4<BR>\n" );
document.write( "	(x+2)^2+(y+3)^2=4<BR>\n" );
document.write( "	(x+2)^2+(y-3)^2=5<BR>\n" );
document.write( "	(x+2)^2+(y+3)^2=5<BR>\n" );
document.write( "	(x+2)^2+(y-3)^2=2\r<BR>\n" );
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document.write( "Can you please explain how you got the answer?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #460196 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5328)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><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/cgi-bin/show-face.mpl?user=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=460196\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-6%2C2%2C-1%2C7%2C%0D%0Agrid%281%29%2C%0D%0Ablue%28circle%28-2%2C3%2C2%29%29%2C%0D%0Ablue%28circle%28-2%2C3%2C0.1%29%29%2C%0D%0Agreen%28line%28-6%2C3%2C2%2C3%29%29%2C+locate%28-5.9%2C3%2Cgreen%28y=3%29%29%2C%0D%0Alocate%28-1.9%2C3%2CO%29%2Clocate%280.1%2C3%2CP%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > The tangent to a circle is perpendicular to the radius at the point of tangency.<BR>\n" );
document.write( "Since the y-axis is vertical, at the point of tangency, rhe radius of a circle tangent to the y-axis must be horizontal.<BR>\n" );
document.write( "The horizontal line that passes through point O(-2,3) is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=3 ALIGN=MIDDLE  ALT=\"y=3\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so<BR>\n" );
document.write( "1) a radius of the circle with center at (-2,3) and tangent to the y-axis is contained in <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=3 ALIGN=MIDDLE  ALT=\"y=3\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and<BR>\n" );
document.write( "2) the point of tangency of the circle with center at (-2,3) and the y-axis is the intersection of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=3 ALIGN=MIDDLE  ALT=\"y=3\"   DISABLED_style= \"max-width:90%; height:auto;\" > and the y-axis.<BR>\n" );
document.write( "That point of tangency is P(0,3).<BR>\n" );
document.write( "The radius at that point is the segment OP, and its length is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0-%28-2%29=2 ALIGN=MIDDLE  ALT=\"0-%28-2%29=2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "The equation of the circle with center at (-2,3) and radius 2 is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%5E2%2B%28y-3%29%5E2=2%5E2 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%5E2%2B%28y-3%29%5E2=2%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > --> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28%28x%2B2%29%5E2%2B%28y-3%29%5E2=4%29 ALIGN=MIDDLE  ALT=\"highlight%28%28x%2B2%29%5E2%2B%28y-3%29%5E2=4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "because the equation of a circle with center at (h,k) and radius r is<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( "but you do not need to remember that formula because<BR>\n" );
document.write( "1) you may know that the distance from a point (x,y) to point O(-2,3) is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28%28x%2B2%29%5E2%2B%28y-3%29%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%28%28x%2B2%29%5E2%2B%28y-3%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > or<BR>\n" );
document.write( "2) you may apply the Pytagorean theorem to the right triangle in the figure below to find that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%5E2%2B%28y-3%29%5E2=2%5E2 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%5E2%2B%28y-3%29%5E2=2%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-0.55%2C0.15%2C-0.05%2C0.65%2C%0D%0Agrid%280%29%2C%0D%0Ablue%28circle%28-0.2%2C0.3%2C0.2%29%29%2C%0D%0Ablue%28circle%28-0.2%2C0.3%2C0.01%29%29%2C%0D%0Ablue%28circle%28-0.1%2C0.473%2C0.01%29%29%2C%0D%0Ablue%28circle%28-0.1%2C0.3%2C0.01%29%29%2C%0D%0Agreen%28line%28-6%2C0.3%2C2%2C0.3%29%29%2Clocate%28-0.54%2C0.3%2Cgreen%28y=3%29%29%2C%0D%0Ablue%28triangle%28-0.19%2C0.3%2C-0.1%2C0.473%2C-0.1%2C0.3%29%29%2C%0D%0Ablue%28rectangle%28-0.1%2C0.3%2C-0.13%2C0.33%29%29%2C%0D%0Alocate%28-0.32%2C0.35%2CO%28-2%2C3%29%29%2C%0D%0Alocate%28-0.15%2C0.53%2CR%28x%2Cy%29%29%2C%0D%0Alocate%28-0.13%2C0.3%2CS%28x%2C3%29%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=system%28OR%5E2=2%5E2%2COS%5E2=%28x%2B3%29%5E2%2CSR%5E2=%28y-3%29%5E2%29 ALIGN=MIDDLE  ALT=\"system%28OR%5E2=2%5E2%2COS%5E2=%28x%2B3%29%5E2%2CSR%5E2=%28y-3%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > --> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B2%29%5E2%2B%28y-3%29%5E2=2%5E2 ALIGN=MIDDLE  ALT=\"%28x%2B2%29%5E2%2B%28y-3%29%5E2=2%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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