Question 253258
{{{4x^2+4y^2-24+16y-100=0}}} Start with the given equation.



{{{4x^2+4y^2+16y-124=0}}} Combine like terms.



{{{4x^2+4y^2+16y=124}}} Add 124 to both sides.



{{{4x^2+(4y^2+16y)=124}}} Group the 'y' terms (ie any terms that have a 'y' in them)



{{{4x^2+4(y^2+4y)=124}}} From that group alone, factor out the GCF 4



Take half of the 'y' coefficient 4 and square it to get {{{(4/2)^2=2^2=4}}}. Add AND subtract this value inside the parenthesis (so the equation isn't changed)



{{{4x^2+4(y^2+4y+highlight(4-4))=124}}} Add AND subtract the value 4 inside the parenthesis



{{{4x^2+4(y^2+4y+4-4)=124}}}



{{{4x^2+4((y^2+4y+4)-4)=124}}} Inside the parenthesis, group the first three terms.



{{{4x^2+4((y+2)^2-4)=124}}} Factor {{{y^2+4y+4}}} (which is a perfect square) to get {{{(y+2)^2}}}



{{{4x^2+4(y+2)^2-16=124}}} Distribute.



{{{4x^2+4(y+2)^2=124+16}}} Add 16 to both sides.



{{{4x^2+4(y+2)^2=140}}} Combine like terms.



{{{4(x^2+(y+2)^2)=140}}} Factor the GCF 4 from the left side.



{{{x^2+(y+2)^2=140/4}}} Divide both sides by 4.



{{{x^2+(y+2)^2=35}}} Reduce.



Take note that we can write {{{x^2}}} as {{{(x-0)^2}}} and {{{(y+2)^2}}} as {{{(y--2)^2}}}. Also, we can write {{{35}}} as {{{(sqrt(35))^2}}}. So the last equation then becomes


{{{(x-0)^2+(y--2)^2=(sqrt(35))^2}}}



which is in the form {{{(x-h)^2+(y-k)^2=r^2}}} (which is a circle) where {{{h=0}}}, {{{k=-2}}}, and {{{r=sqrt(35)}}}. Recall that (h,k) is the center. So the center is (0,-2). Also, the radius is 'r' which means that the radius is {{{sqrt(35)}}} units.



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Answer:


So the standard form of the equation {{{4x^2+4y^2-24+16y-100=0}}} is {{{(x-0)^2+(y--2)^2=(sqrt(35))^2}}} which is a circle with radius of {{{sqrt(35)}}} units and a center at (0,-2).



Note: your teacher will probably want a simplified answer. So s/he will probably want the answer of {{{x^2+(y+2)^2=35}}} (since it's much cleaner looking). If you want to verify the answer, simply graph the two conic sections and you'll find that they are the same conic section.