Question

rewrite from the general form to center - radius form. then label center, radius and graph
( x^2 + y^2 + 2x - 4y - 11 = 0 )

1. ( x^2 + y^2 + 10x - 6y + 30 = 0 )
2. ( x^2 + y^2 - 8x + 10y + 5 = 0 )
3. ( x^2 + y^2 + 12x - 10y - 30 = 0 )

write the equation of each circle in center - radius form.
(grids for graphing are shown)

Explanation:

Let's take problem 11: \(x^{2}+y^{2}+10x - 6y+30 = 0\) and convert it to center - radius form \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Step 1: Group \(x\) - terms and \(y\) - terms together

We rewrite the equation as \((x^{2}+10x)+(y^{2}-6y)=- 30\)

Step 2: Complete the square for \(x\) - terms

For the \(x\) - terms \(x^{2}+10x\), we use the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\). Here, \(a = x\) and \(2ab=10x\), so \(b = 5\) (since \(2x\times5 = 10x\)). Then \(x^{2}+10x=(x + 5)^{2}-25\)

Step 3: Complete the square for \(y\) - terms

For the \(y\) - terms \(y^{2}-6y\), using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\). Here, \(a = y\) and \(2ab = 6y\), so \(b = 3\) (since \(2y\times3=6y\)). Then \(y^{2}-6y=(y - 3)^{2}-9\)

Step 4: Substitute the completed squares back into the equation

Substitute \((x + 5)^{2}-25\) for \(x^{2}+10x\) and \((y - 3)^{2}-9\) for \(y^{2}-6y\) in the equation \((x^{2}+10x)+(y^{2}-6y)=-30\)

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

Step 5: Simplify the equation

First, combine the constant terms on the left - hand side: \((x + 5)^{2}+(y - 3)^{2}-34=-30\)

Then, add 34 to both sides of the equation: \((x + 5)^{2}+(y - 3)^{2}=-30 + 34\)

So, \((x + 5)^{2}+(y - 3)^{2}=4\) or \((x + 5)^{2}+(y - 3)^{2}=2^{2}\)

The center of the circle is \((-5,3)\) and the radius \(r = 2\)

Let's take problem 12: \(x^{2}+y^{2}-8x + 10y+5 = 0\)

Step 1: Group \(x\) - terms and \(y\) - terms together

Rewrite the equation as \((x^{2}-8x)+(y^{2}+10y)=-5\)

Step 2: Complete the square for \(x\) - terms

For \(x^{2}-8x\), using \((a - b)^{2}=a^{2}-2ab + b^{2}\), where \(a=x\) and \(2ab = 8x\), so \(b = 4\) (since \(2x\times4 = 8x\)). Then \(x^{2}-8x=(x - 4)^{2}-16\)

Step 3: Complete the square for \(y\) - terms

For \(y^{2}+10y\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2ab=10y\), so \(b = 5\) (since \(2y\times5=10y\)). Then \(y^{2}+10y=(y + 5)^{2}-25\)

Step 4: Substitute the completed squares back into the equation

Substitute \((x - 4)^{2}-16\) for \(x^{2}-8x\) and \((y + 5)^{2}-25\) for \(y^{2}+10y\) in \((x^{2}-8x)+(y^{2}+10y)=-5\)

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

Step 5: Simplify the equation

Combine the constant terms: \((x - 4)^{2}+(y + 5)^{2}-41=-5\)

Add 41 to both sides: \((x - 4)^{2}+(y + 5)^{2}=-5 + 41\)

So, \((x - 4)^{2}+(y + 5)^{2}=36\) or \((x - 4)^{2}+(y + 5)^{2}=6^{2}\)

The center is \((4,-5)\) and the radius \(r = 6\)

Let's take problem 13: \(x^{2}+y^{2}+12x-10y - 30=0\)

Step 1: Group \(x\) - terms and \(y\) - terms together

Rewrite the equation as \((x^{2}+12x)+(y^{2}-10y)=30\)

Step 2: Complete the square for \(x\) - terms

For \(x^{2}+12x\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a=x\) and \(2ab = 12x\), so \(b = 6\) (since \(2x\times6=12x\)). Then \(x^{2}+12x=(x + 6)^{2}-36\)

Step 3: Complete the square for \(y\) - terms

For \(y^{2}-10y\), using \((a - b)^{2}=a^{2}-2ab + b^{2}\), where \(a = y\) and \(2ab=10y\), so \(b = 5\) (since \(2y\times5 = 10y\)). Then \(y^{2}-10y=(y - 5)^{2}-25\)

Step 4: Substitute the completed squares back into the equation

Substitute \((x + 6)^{2}-36\) for \(x^{2}+12x\) and \((y - 5)^{2}-25\) for \(y^{2}-10y\) in \((x^{2}+12x)+(y^{2}-10y)=30\)

We get \((x + 6)^{2}-36+(y - 5)^{2}-25=30\)

Step 5: Simplify the equation

Combine the constant terms: \((x + 6)^{2}+(y - 5)^{2}-61=30\)

Add 61 to both sides: \((x + 6)^{2}+(y - 5)^{2}=30 + 61\)

So, \((x + 6)^{2}+(y - 5)^{2}=91\) or \((x + 6)^{2}+(y - 5)^{2}=(\sqrt{91})^{2}\)

T…

Step1: Group x and y terms

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

Step2: Complete square for x

\(x^{2}+2x=(x + 1)^{2}-1\)

Step3: Complete square for y

\(y^{2}-4y=(y - 2)^{2}-4\)

Step4: Substitute and simplify

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

Answer:

Center - radius form: \(\boldsymbol{(x + 1)^{2}+(y - 2)^{2}=16}\), Center: \(\boldsymbol{(-1,2)}\), Radius: \(\boldsymbol{4}\)