Question

now, rewrite from the general form to center - radius form. then, label center, radius, and graph.

1. ( x^{2}+y^{2}+2x - 4y - 11 = 0 )
2. ( x^{2}+y^{2}+10x - 6y + 30 = 0 )

write the equation of each circle in center - radius form.

1. ( x^{2}+y^{2}-8x + 10y + 5 = 0 )
2. ( x^{2}+y^{2}+12x - 10y - 20 = 0 )

Explanation:

Step1: Group \(x\) and \(y\) terms

For \(x^{2}+y^{2}+2x - 4y-11 = 0\), we have \((x^{2}+2x)+(y^{2}-4y)=11\)

Step2: Complete the square for \(x\) terms

For \(x^{2}+2x\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 2x\Rightarrow b = 1\). So \(x^{2}+2x=(x + 1)^{2}-1\)

Step3: Complete the square for \(y\) terms

For \(y^{2}-4y\), using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = y\), \(2ab=4y\Rightarrow b = 2\). So \(y^{2}-4y=(y - 2)^{2}-4\)

Step4: Substitute back into the equation

\((x + 1)^{2}-1+(y - 2)^{2}-4=11\)
\((x + 1)^{2}+(y - 2)^{2}=11 + 1+4\)
\((x + 1)^{2}+(y - 2)^{2}=16\)

Center \((-1,2)\), radius \(r = 4\)

For \(x^{2}+y^{2}+10x-6y + 30=0\)

Step1: Group \(x\) and \(y\) terms

\((x^{2}+10x)+(y^{2}-6y)=- 30\)

Step2: Complete the square for \(x\) terms

For \(x^{2}+10x\), \(a=x\), \(2ab = 10x\Rightarrow b = 5\). So \(x^{2}+10x=(x + 5)^{2}-25\)

Step3: Complete the square for \(y\) terms

For \(y^{2}-6y\), \(a = y\), \(2ab=6y\Rightarrow b = 3\). So \(y^{2}-6y=(y - 3)^{2}-9\)

Step4: Substitute back into the equation

\((x + 5)^{2}-25+(y - 3)^{2}-9=-30\)
\((x + 5)^{2}+(y - 3)^{2}=-30+25 + 9\)
\((x + 5)^{2}+(y - 3)^{2}=4\)

Center \((-5,3)\), radius \(r = 2\)

For \(x^{2}+y^{2}-8x + 10y+5=0\)

Step1: Group \(x\) and \(y\) terms

\((x^{2}-8x)+(y^{2}+10y)=-5\)

Step2: Complete the square for \(x\) terms

For \(x^{2}-8x\), \(a=x\), \(2ab = 8x\Rightarrow b = 4\). So \(x^{2}-8x=(x - 4)^{2}-16\)

Step3: Complete the square for \(y\) terms

For \(y^{2}+10y\), \(a = y\), \(2ab=10y\Rightarrow b = 5\). So \(y^{2}+10y=(y + 5)^{2}-25\)

Step4: Substitute back into the equation

\((x - 4)^{2}-16+(y + 5)^{2}-25=-5\)
\((x - 4)^{2}+(y + 5)^{2}=-5+16 + 25\)
\((x - 4)^{2}+(y + 5)^{2}=36\)

For \(x^{2}+y^{2}+12x-10y-20=0\)

Step1: Group \(x\) and \(y\) terms

\((x^{2}+12x)+(y^{2}-10y)=20\)

Step2: Complete the square for \(x\) terms

For \(x^{2}+12x\), \(a=x\), \(2ab = 12x\Rightarrow b = 6\). So \(x^{2}+12x=(x + 6)^{2}-36\)

Step3: Complete the square for \(y\) terms

For \(y^{2}-10y\), \(a = y\), \(2ab=10y\Rightarrow b = 5\). So \(y^{2}-10y=(y - 5)^{2}-25\)

Step4: Substitute back into the equation

\((x + 6)^{2}-36+(y - 5)^{2}-25=20\)
\((x + 6)^{2}+(y - 5)^{2}=20+36 + 25\)
\((x + 6)^{2}+(y - 5)^{2}=81\)

Answer:

1. Center - radius form: \((x + 1)^{2}+(y - 2)^{2}=16\), Center \((-1,2)\), radius \(r = 4\)
2. Center - radius form: \((x + 5)^{2}+(y - 3)^{2}=4\), Center \((-5,3)\), radius \(r = 2\)
3. Center - radius form: \((x - 4)^{2}+(y + 5)^{2}=36\)
4. Center - radius form: \((x + 6)^{2}+(y - 5)^{2}=81\)