Question

Graph the circle (2x - 3)2 + (y + 3)2 =36

Answer 1

Answer:

$\mathrm{Ellipse\:with\:center}\:\left(h,\:k\right)=\left(\frac{3}{2},\:-3\right),\:\:\mathrm{semi-major\:axis}\:b=6,\:\:\mathrm{semi-minor\:axis}\:a=3$

Step-by-step explanation:

$\left(2x-3\right)^2+\left(y+3\right)^2=36\\\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{is\:the\:ellipse\:standard\:equation}\\\mathrm{with\:center}\:\left(h,\:k\right)\:\mathrm{and\:}a,\:b\mathrm{\:are\:the\:semi-major\:and\:semi-minor\:axes}\\\mathrm{Rewrite}\:\left(2x-3\right)^2+\left(y+3\right)^2=36\:\mathrm{in\:the\:form\:of\:the\:standard\:ellipse\:equation}\\\left(2x-3\right)^2+\left(y+3\right)^2=36\\\mathrm{Rewrite\:as}\\\left(2x-3\right)^2+\left(y+3\right)^2-36=0$

$\mathrm{Simplify}\:\left(2x-3\right)^2+\left(y+3\right)^2-36:\quad 4x^2-12x+y^2+6y-18\\4x^2-12x+y^2+6y-18=0\\\mathrm{Add\:}18\mathrm{\:to\:both\:sides}\\4x^2-12x+y^2+6y=18\\Factor\:out\:coefficient\:of\:square\:terms\\4\left(x^2-3x\right)+\left(y^2+6y\right)=18\\\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4\\\left(x^2-3x\right)+\frac{1}{4}\left(y^2+6y\right)=\frac{9}{2}\\\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1$

$\frac{1}{1}\left(x^2-3x\right)+\frac{1}{4}\left(y^2+6y\right)=\frac{9}{2}\\\mathrm{Convert}\:x\:\mathrm{to\:square\:form}\\\frac{1}{1}\left(x^2-3x+\frac{9}{4}\right)+\frac{1}{4}\left(y^2+6y\right)=\frac{9}{2}+\frac{1}{1}\left(\frac{9}{4}\right)\\\mathrm{Convert\:to\:square\:form}\\\frac{1}{1}\left(x-\frac{3}{2}\right)^2+\frac{1}{4}\left(y^2+6y\right)=\frac{9}{2}+\frac{1}{1}\left(\frac{9}{4}\right)\\\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$

$\frac{1}{1}\left(x-\frac{3}{2}\right)^2+\frac{1}{4}\left(y^2+6y+9\right)=\frac{9}{2}+\frac{1}{1}\left(\frac{9}{4}\right)+\frac{1}{4}\left(9\right)\\\mathrm{Convert\:to\:square\:form}\\\frac{1}{1}\left(x-\frac{3}{2}\right)^2+\frac{1}{4}\left(y+3\right)^2=\frac{9}{2}+\frac{1}{1}\left(\frac{9}{4}\right)+\frac{1}{4}\left(9\right)\\\mathrm{Refine\:}\frac{9}{2}+\frac{1}{1}\left(\frac{9}{4}\right)+\frac{1}{4}\left(9\right)\\\frac{1}{1}\left(x-\frac{3}{2}\right)^2+\frac{1}{4}\left(y+3\right)^2=9$

$\mathrm{Divide\:by}\:9\\\frac{\left(x-\frac{3}{2}\right)^2}{9}+\frac{\left(y+3\right)^2}{36}=1\\\mathrm{Rewrite\:in\:standard\:form}\\\frac{\left(x-\frac{3}{2}\right)^2}{3^2}+\frac{\left(y-\left(-3\right)\right)^2}{6^2}=1\\\mathrm{Therefore\:ellipse\:properties\:are:}\\\left(h,\:k\right)=\left(\frac{3}{2},\:-3\right),\:a=3,\:b=6\\b>a\:\mathrm{therefore}\:b\:\mathrm{is\:semi-major\:axis\:and}\:a\:\mathrm{is\:semi-minor\:axis}$

$\mathrm{Ellipse\:with\:center}\:\left(h,\:k\right)=\left(\frac{3}{2},\:-3\right),\:\:\mathrm{semi-major\:axis}\:b=6,\:\:\mathrm{semi-minor\:axis}\:a=3$

Answer 2

Answer: 9 on the x axis  and 18 on the y axis