document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1082996>Question 1082996</A>: <I><SPAN STYLE=\"word-break: break-all;\"> C is a circle centered at (1,-2) with radius 3. Q is a quadratic function with the same x-intercepts as C and whose vertex is on C.<BR>\n" );
document.write( " (a) write an equation for C and find it’s x-intercepts.<BR>\n" );
document.write( " (b) Find All Possible equations for the quad Q and explain how you know that the equations you have given are the only ones that will work </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #696999 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=696999\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The circle is (x-1)^2+(Y+2)^2=9<BR>\n" );
document.write( "the quadratic function has the same x-intercepts.<BR>\n" );
document.write( "The x-intercepts of the circle are where y=0, and the equation is x^2-2x+1+4=9<BR>\n" );
document.write( "x^2-2x-4=0<BR>\n" );
document.write( "x=(1/2)(2+/-sqrt(4+16)=(1/2)(2+/- 2 sqrt (5)=1+/-sqrt (5)<BR>\n" );
document.write( "If the vertex is on the circle, it has to be symmetrical with the x-value of the center, so that x=1<BR>\n" );
document.write( "But the vertex x-value equals -b/2a, so -b/2a=1 and b=-2a<BR>\n" );
document.write( "Therefore, ax^2+bx+c=0 and ax^2-2ax+c=0<BR>\n" );
document.write( "substituting the values of the roots for x<BR>\n" );
document.write( "a(1+ sqrt(5)^2-2a(1+sqrt(5))+c=0<BR>\n" );
document.write( "a(6+2 sqrt(5))-2a(1- sqrt(5))+c=0<BR>\n" );
document.write( "6a +2a sqrt(5)-2a+2a sqrt(5)+c=0<BR>\n" );
document.write( "4a+c=0<BR>\n" );
document.write( "c=-4a<BR>\n" );
document.write( "and ax^2+bx+c becomes ax^2-2ax-4a=0, which is a(x^2-2x-4)=0 which is the above.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-2%2Bsqrt%28-%28x-1%29%5E2%2B9%29%2C-2-sqrt%28-%28x-1%29%5E2%2B9%29%2Cx%5E2-2x-4%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "There is another that has a negative leading coefficient, and that has an intercept -1/5 of the other, so it is the same quadratic with a=-0.2.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-2%2Bsqrt%28-%28x-1%29%5E2%2B9%29%2C-2-sqrt%28-%28x-1%29%5E2%2B9%29%2Cx%5E2-2x-4%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-2%2Bsqrt%28-%28x-1%29%5E2%2B9%29%2C-2-sqrt%28-%28x-1%29%5E2%2B9%29%2C-%28.2%29%28x%5E2-2x-4%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "There can't be other equations, because they would have a vertex on another part of the circle and for the roots to match, the vertex has to be at x=1.<BR>\n" );
document.write( " </SPAN>\n" );
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