91Ó°ÊÓ

Completing the square is a technique used to convert a quadratic expression into a perfect square trinomial. This is often done to simplify solving quadratic equations or to rewrite them in a more useful form, such as the standard form of a circle equation.
In our exercise, we had the equation \( x^2 + 7x + y^2 = 2 \). *Completing the square* helps to handle the \( x \) terms effectively, making it easier to break them into a single square term.
To complete the square for the \( x \) terms, we follow these steps:

• Take the coefficient of \( x \), which is 7, and halve it to get \( \frac{7}{2} \).
• Then square this result: \( \left(\frac{7}{2}\right)^2 = \frac{49}{4} \).
• Add and subtract \( \frac{49}{4} \) inside the equation: \( x^2 + 7x \) becomes \( \left(x + \frac{7}{2}\right)^2 - \frac{49}{4} \).

By doing this, the equation turns into a complete square form, which is much easier for identifying the circle's properties later.

The standard form of a circle's equation helps quickly identify the circle's center and radius. It's expressed as \[(x-h)^2 + (y-k)^2 = r^2\]. Here, \( h \) and \( k \) are the coordinates of the center, and \( r \) is the radius.
In this format, both \( x \) and \( y \) terms form a perfect square, which makes calculating the circle's characteristics straightforward.
After completing the square in the previous example, the circle equation was reorganized into the standard form: \[ \left(x + \frac{7}{2}\right)^2 + y^2 = \frac{57}{4} \].- Here, it becomes clear that \( \left(x + \frac{7}{2}\right)^2 \) and \( y^2 \) map to the squares needed for the standard circle form.
These notations make set-up, breakdown, and result interpretation in complex equations more accessible for learners.

Determining a circle's center and radius from its equation is crucial in understanding a circle in geometry. Using the standard form \[(x-h)^2 + (y-k)^2 = r^2\], we can easily extract these properties.
From our exercise's equation: \[ \left(x + \frac{7}{2}\right)^2 + y^2 = \frac{57}{4} \].- The circle center \( (h, k) \) is directly identified from \((x + h)\) and \((y + k)\). Here, our equation implies,

• \( h = -\frac{7}{2} \) and \( k = 0 \), making the center \( \left( -\frac{7}{2}, 0 \right) \).
• For the radius \( r \), observe \( r^2 = \frac{57}{4} \). Thus, \( r = \sqrt{\frac{57}{4}} \), simplified to \( \frac{\sqrt{57}}{2} \).

Calculating these values clarifies the circle's geometrical positioning and its expanse in the coordinate system, enhancing understanding and proficiency with circles in mathematics.