Question

match the circle equations in general form with their corresponding equations in standard form

Answer:

To solve this, we convert each general - form circle equation to standard form \((x - h)^2+(y - k)^2=r^2\) by completing the square.

1. For the equation \(x^{2}+y^{2}-4x + 12y-20 = 0\)

• Step 1: Group \(x\) and \(y\) terms
• Rearrange the equation as \((x^{2}-4x)+(y^{2}+12y)=20\).
• Step 2: Complete the square for \(x\) and \(y\) terms
• For the \(x\) - terms: \(x^{2}-4x=(x - 2)^{2}-4\) (using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\) and \(2ab = 4x\), so \(b = 2\)).
• For the \(y\) - terms: \(y^{2}+12y=(y + 6)^{2}-36\) (using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\) and \(2ab = 12y\), so \(b = 6\)).
• Substitute these back into the equation: \((x - 2)^{2}-4+(y + 6)^{2}-36 = 20\).
• Simplify: \((x - 2)^{2}+(y + 6)^{2}=20 + 4+36=60\).

So, \(x^{2}+y^{2}-4x + 12y-20 = 0\) matches \((x - 2)^{2}+(y + 6)^{2}=60\).

2. For the equation \(x^{2}+y^{2}+6x-8y - 10 = 0\)

• Step 1: Group \(x\) and \(y\) terms
• Rearrange the equation as \((x^{2}+6x)+(y^{2}-8y)=10\).
• Step 2: Complete the square for \(x\) and \(y\) terms
• For the \(x\) - terms: \(x^{2}+6x=(x + 3)^{2}-9\) (using \((a + b)^2=a^{2}+2ab + b^{2}\), \(a=x\), \(2ab = 6x\), \(b = 3\)).
• For the \(y\) - terms: \(y^{2}-8y=(y - 4)^{2}-16\) (using \((a - b)^2=a^{2}-2ab + b^{2}\), \(a = y\), \(2ab = 8y\), \(b = 4\)).
• Substitute back: \((x + 3)^{2}-9+(y - 4)^{2}-16 = 10\).
• Simplify: \((x + 3)^{2}+(y - 4)^{2}=10 + 9+16=35\). Wait, there is a mistake. Wait, let's re - check. Wait, the right - hand side standard form we have is \((x + 1)^{2}+(y - 6)^{2}=46\) and others. Wait, let's re - do this.
• Wait, the correct completion:
• \(x^{2}+6x=(x + 3)^{2}-9\), \(y^{2}-8y=(y - 4)^{2}-16\)
• The equation becomes \((x + 3)^{2}-9+(y - 4)^{2}-16=10\)
• \((x + 3)^{2}+(y - 4)^{2}=10 + 9+16 = 35\). But this is not in our list. Wait, maybe I made a mistake. Wait, the other standard form is \((x + 1)^{2}+(y - 6)^{2}=46\) and the general form \(x^{2}+y^{2}+2x-12y - 9 = 0\)
• Let's take \(x^{2}+y^{2}+2x-12y - 9 = 0\)
• Group: \((x^{2}+2x)+(y^{2}-12y)=9\)
• Complete the square: \(x^{2}+2x=(x + 1)^{2}-1\), \(y^{2}-12y=(y - 6)^{2}-36\)
• Substitute: \((x + 1)^{2}-1+(y - 6)^{2}-36=9\)
• \((x + 1)^{2}+(y - 6)^{2}=9 + 1+36=46\)
• And the general form \(x^{2}+y^{2}+6x-8y - 10 = 0\)
• Group: \((x^{2}+6x)+(y^{2}-8y)=10\)
• \(x^{2}+6x=(x + 3)^{2}-9\), \(y^{2}-8y=(y - 4)^{2}-16\)
• \((x + 3)^{2}-9+(y - 4)^{2}-16=10\)
• \((x + 3)^{2}+(y - 4)^{2}=35\). This is wrong. Wait, maybe the original problem has a typo or I misread. Wait, let's take the third equation: \(3x^{2}+3y^{2}+12x + 18y-15 = 0\)

3. For the equation \(3x^{2}+3y^{2}+12x + 18y-15 = 0\)

• Step 1: Divide the entire equation by 3
• We get \(x^{2}+y^{2}+4x + 6y-5 = 0\).
• Step 2: Group \(x\) and \(y\) terms
• \((x^{2}+4x)+(y^{2}+6y)=5\).
• Step 3: Complete the square
• For \(x\) - terms: \(x^{2}+4x=(x + 2)^{2}-4\) (since \(2ab = 4x\), \(a=x\), \(b = 2\)).
• For \(y\) - terms: \(y^{2}+6y=(y + 3)^{2}-9\) (since \(2ab = 6y\), \(a = y\), \(b = 3\)).
• Substitute back: \((x + 2)^{2}-4+(y + 3)^{2}-9 = 5\).
• Simplify: \((x + 2)^{2}+(y + 3)^{2}=5 + 4+9=18\).

So, \(3x^{2}+3y^{2}+12x + 18y-15 = 0\) matches \((x + 2)^{2}+(y + 3)^{2}=18\).

4. For the equation \(5x^{2}+5y^{2}-10x + 20y-30 = 0\)

• Step 1: Divide the entire equation by 5
• We get \(x^{2}+y^{2}-2x + 4y-6 = 0\).
• Step 2: Group \(x\) and \(y\) terms
• \((x^{2}-2x)+(y^{2}+4y)=6\).
• Step 3: Complete the square
• For \(x\) - terms: \(x^{2}-2x=(x - 1)^{2}-1\) (since \(2ab = 2x\), \(a=x\), \(b = 1\)).
• For \(y\) - terms: \(y^{2}+4y=(y + 2)^{2}-4\) (since \(2ab = 4y\), \(a = y\), \(b = 2\)).
• Substitute back: \((x - 1)^{2}-1+(y + 2)^{2}-4 = 6\).
• Simplify: \((x - 1)^{2}+(y + 2)^{2}=6 + 1+4=11\). This is not in our list. Wait, maybe I made a mistake. Wait, let's take the fifth equation: \(2x^{2}+2y^{2}-24x-16y - 8 = 0\)

5. For the equation \(2x^{2}+2y^{2}-24x-16y - 8 = 0\)

• Step 1: Divide the entire equation by 2
• We get \(x^{2}+y^{2}-12x-8y - 4 = 0\).
• Step 2: Group \(x\) and \(y\) terms
• \((x^{2}-12x)+(y^{2}-8y)=4\).
• Step 3: Complete the square
• For \(x\) - terms: \(x^{2}-12x=(x - 6)^{2}-36\) (since \(2ab = 12x\), \(a=x\), \(b = 6\)).
• For \(y\) - terms: \(y^{2}-8y=(y - 4)^{2}-16\) (since \(2ab = 8y\), \(a = y\), \(b = 4\)).
• Substitute back: \((x - 6)^{2}-36+(y - 4)^{2}-16 = 4\).
• Simplify: \((x - 6)^{2}+(y - 4)^{2}=4 + 36+16=56\).

So, \(2x^{2}+2y^{2}-24x-16y - 8 = 0\) matches \((x - 6)^{2}+(y - 4)^{2}=56\).

6. For the equation \(x^{2}+y^{2}+2x-12y - 9 = 0\)

• Step 1: Group \(x\) and \(y\) terms
• \((x^{2}+2x)+(y^{2}-12y)=9\).
• Step 2: Complete the square
• For \(x\) - terms: \(x^{2}+2x=(x + 1)^{2}-1\) (since \(2ab = 2x\), \(a=x\), \(b = 1\)).
• For \(y\) - terms: \(y^{2}-12y=(y - 6)^{2}-36\) (since \(2ab = 12y\), \(a = y\), \(b = 6\)).
• Substitute back: \((x + 1)^{2}-1+(y - 6)^{2}-36 = 9\).
• Simplify: \((x + 1)^{2}+(y - 6)^{2}=9 + 1+36=46\).

And the general form \(x^{2}+y^{2}+6x-8y - 10 = 0\)

• Group: \((x^{2}+6x)+(y^{2}-8y)=10\)
• \(x^{2}+6x=(x + 3)^{2}-9\), \(y^{2}-8y=(y - 4)^{2}-16\)
• \((x + 3)^{2}-9+(y - 4)^{2}-16=10\)
• \((x + 3)^{2}+(y - 4)^{2}=35\). This is not in our list. Wait, maybe the original problem has a pairing error or I misread. But from the above correct completions:

• \(x^{2}+y^{2}-4x + 12y-20 = 0\) \(\longleftrightarrow\) \((x - 2)^{2}+(y + 6)^{2}=60\)
• \(3x^{2}+3y^{2}+12x + 18y-15 = 0\) \(\longleftrightarrow\) \((x + 2)^{2}+(y + 3)^{2}=18\)
• \(2x^{2}+2y^{2}-24x-16y - 8 = 0\) \(\longleftrightarrow\) \((x - 6)^{2}+(y - 4)^{2}=56\)
• \(x^{2}+y^{2}+2x-12y - 9 = 0\) \(\longleftrightarrow\) \((x + 1)^{2}+(y - 6)^{2}=46\)
• \(x^{2}+y^{2}+6x-8y - 10 = 0\) Let's re - check. Wait, if we take \(x^{2}+y^{2}+6x-8y - 10 = 0\) and try to match with the remaining. Wait, maybe I made a mistake in the first equation. Wait, the first general form \(x^{2}+y^{2}-4x + 12y-20 = 0\)
• \(x^{2}-4x=(x - 2)^{2}-4\), \(y^{2}+12y=(y + 6)^{2}-36\)
• \((x - 2)^{2}-4+(y + 6)^{2}-36=20\)
• \((x - 2)^{2}+(y + 6)^{2}=60\) (correct)

The second general form \(x^{2}+y^{2}+6x-8y - 10 = 0\)

• \(x^{2}+6x=(x + 3)^{2}-9\), \(y^{2}-8y=(y - 4)^{2}-16\)
• \((x + 3)^{2}-9+(y - 4)^{2}-16=10\)
• \((x + 3)^{2}+(y - 4)^{2}=35\). But there is no standard form with 35. So, maybe the intended pairing is:

General Form	Standard Form
\(x^{2}+y^{2}+6x-8y - 10 = 0\)	Let's check the standard form \((x + 1)^{2}+(y - 6)^{2}=46\) is for \(x^{2}+y^{2}+2x-12y - 9 = 0\)
\(3x^{2}+3y^{2}+12x + 18y-15 = 0\)	\((x + 2)^{2}+(y + 3)^{2}=18\)
\(5x^{2}+5y^{2}-10x + 20y-30 = 0\)	Let's divide by 5: \(x^{2}+y^{2}-2x + 4y-6 = 0\), group: \((x - 1)^{2}-1+(y + 2)^{2}-4=6\), \((x - 1)^{2}+(y + 2)^{2}=11\) (no match)
\(2x^{2}+2y^{2}-24x-16y - 8 = 0\)	\((x - 6)^{2}+(y - 4)^{2}=56\)
\(x^{2}+y^{2}+2x-12y - 9 = 0\)	\((x + 1)^{2}+(y - 6)^{2}=46\)

The correct pairings are:

• \(x^{2}+y^{2}-4x + 12y-20 = 0\) \(\longleftrightarrow\) \((x - 2)^{2}+(y + 6)^{2}=60\)
• \(3x^{2}+3y^{2}+12x + 18y-15 = 0\) \(\longleftrightarrow\) \((x + 2)^{2}+(y + 3)^{2}=18\)
• \(2x^{2}+2y^{2}-24x-16y - 8 = 0\) \(\longleftrightarrow\) \((x - 6)^{2}+(y - 4)^{2}=56\)
• \(x^{2}+y^{2}+2x-12y - 9 = 0\) \(\longleftrightarrow\) \((x + 1)^{2}+(y - 6)^{2}=46\)
• \(x^{2}+y^{2}+6x-8y - 10 = 0\) (There may be an error in