Verified Answer :
- Convert the general equation \(x^2 + y^2 + 4x - 6y - 36 = 0\) to standard form by completing the square for both \(x\) and \(y\).
- Simplify the equation to \((x+2)^2 + (y-3)^2 = 49\).
- Identify the center of the circle as \((-2, 3)\) and the radius as \(7\).
- Verify the accuracy of each statement given, finding that only the statements about adding 4 and the location of the center are correct.
Explanation
- Standard form conversion and analysis
To convert the circle equation \(x^2 + y^2 + 4x - 6y - 36 = 0\) into standard form \((x-h)^2 + (y-k)^2 = r^2\), we start by rearranging the terms and isolating the variables:
\[x^2 + 4x + y^2 - 6y = 36\]
To complete the square for the \(x\) terms \(x^2 + 4x\), we take half of the coefficient of \(x\), which is \(\frac{4}{2} = 2\), and square it: \(2^2 = 4\). Adding 4 to both sides gives:
\[x^2 + 4x + 4 + y^2 - 6y = 36 + 4\]
Next, we complete the square for the \(y\) terms \(y^2 - 6y\). We take half of the coefficient of \(y\), which is \(\frac{-6}{2} = -3\), and square it: \((-3)^2 = 9\). Adding 9 to both sides gives:
\[x^2 + 4x + 4 + y^2 - 6y + 9 = 36 + 4 + 9\]
Factoring the expressions, we get:
\[(x+2)^2 + (y-3)^2 = 49\]
Comparing this with the standard form \((x-h)^2 + (y-k)^2 = r^2\), we identify the center \((h, k) = (-2, 3)\) and the radius squared \(r^2 = 49\), so the radius is \(r = \sqrt{49} = 7\).
- Evaluation of the true statements
Based on the analysis, we evaluate the given statements:
- "To begin converting the equation to standard form, subtract \(36\) from both sides." - False, we add 36 to both sides.
- "To complete the square for the \(x\) terms, add \(4\) to both sides." - True, as shown in the derivation.
- "The center of the circle is at \((-2, 3)\)." - True, identified from the standard form.
- "The center of the circle is at \((4, -6)\)." - False.
- "The radius of the circle is \(6\) units." - False, it is 7.
- "The radius of the circle is \(49\) units." - False, 49 is the radius squared.
The true statements are: "To complete the square for the \(x\) terms, add \(4\) to both sides" and "The center of the circle is at \((-2, 3)\)".
Examples
Standard forms of equations are extremely useful in fields like computer graphics and game development. For instance, if you are programming a collision detection system for a circular character moving across a grid, you will need to know the center \((h, k)\) and the radius \(r\) to easily calculate the distance between your character's position and obstacles. By converting general equations into standard forms, you simplify these distance calculations immensely, ensuring your character moves smoothly and interacts correctly with the environment!