Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
The given equation of the circle is [tex]\(x^2 + y^2 + Cx + Dy + E = 0\)[/tex]. This is in the general form of a circle's equation. To understand how changes in the radius affect the coefficients [tex]\(C\)[/tex], [tex]\(D\)[/tex], and [tex]\(E\)[/tex], we can convert this into the standard form of a circle's equation.
### Step 1: Conversion to Standard Form
The standard form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius of the circle.
### Step 2: Deriving the Center and Radius
We can relate the general form to the standard form by completing the square. Starting with the general form:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0 \][/tex]
We complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + Cx + y^2 + Dy + E = 0 \][/tex]
For the [tex]\(x\)[/tex] terms:
[tex]\[ x^2 + Cx \quad \text{can be written as} \quad \left(x + \frac{C}{2}\right)^2 - \left(\frac{C}{2}\right)^2 \][/tex]
For the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 + Dy \quad \text{can be written as} \quad \left(y + \frac{D}{2}\right)^2 - \left(\frac{D}{2}\right)^2 \][/tex]
Thus, rewriting the equation, we get:
[tex]\[ \left(x + \frac{C}{2}\right)^2 - \left(\frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 - \left(\frac{D}{2}\right)^2 + E = 0 \][/tex]
We combine the constant terms:
[tex]\[ \left(x + \frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 - \left(\frac{C}{2}\right)^2 - \left(\frac{D}{2}\right)^2 + E = 0 \][/tex]
[tex]\[ \left(x + \frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 = \left(\frac{C}{2}\right)^2 + \left(\frac{D}{2}\right)^2 - E \][/tex]
This gives us the standard form:
[tex]\[ \left(x + \frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 = r^2 \][/tex]
where:
[tex]\[ h = -\frac{C}{2}, \quad k = -\frac{D}{2}, \quad \text{and} \quad r^2 = \left(\frac{C}{2}\right)^2 + \left(\frac{D}{2}\right)^2 - E \][/tex]
### Step 3: Effect of Decreasing the Radius
If the radius [tex]\(r\)[/tex] is decreased, then [tex]\(r^2\)[/tex] becomes smaller. Given the equation:
[tex]\[ r^2 = \left(\frac{C}{2}\right)^2 + \left(\frac{D}{2}\right)^2 - E \][/tex]
we see that the terms [tex]\(\left(\frac{C}{2}\right)^2\)[/tex] and [tex]\(\left(\frac{D}{2}\right)^2\)[/tex] depend on [tex]\(C\)[/tex] and [tex]\(D\)[/tex], which are functions of the center coordinates. These coordinates are not changing, so [tex]\(C\)[/tex] and [tex]\(D\)[/tex] remain unchanged.
To decrease [tex]\(r^2\)[/tex], the term [tex]\(- E\)[/tex] must become larger (i.e., [tex]\(E\)[/tex] must increase since [tex]\( E\)[/tex] is subtracted).
### Conclusion
Therefore, the coefficients [tex]\(C\)[/tex] and [tex]\(D\)[/tex] are unchanged, but the coefficient [tex]\(E\)[/tex] increases when the radius is decreased.
The correct answer is:
[tex]\[ \boxed{E} \text{: $C$ and $D$ are unchanged, but $E$ increases.} \][/tex]
### Step 1: Conversion to Standard Form
The standard form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius of the circle.
### Step 2: Deriving the Center and Radius
We can relate the general form to the standard form by completing the square. Starting with the general form:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0 \][/tex]
We complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + Cx + y^2 + Dy + E = 0 \][/tex]
For the [tex]\(x\)[/tex] terms:
[tex]\[ x^2 + Cx \quad \text{can be written as} \quad \left(x + \frac{C}{2}\right)^2 - \left(\frac{C}{2}\right)^2 \][/tex]
For the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 + Dy \quad \text{can be written as} \quad \left(y + \frac{D}{2}\right)^2 - \left(\frac{D}{2}\right)^2 \][/tex]
Thus, rewriting the equation, we get:
[tex]\[ \left(x + \frac{C}{2}\right)^2 - \left(\frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 - \left(\frac{D}{2}\right)^2 + E = 0 \][/tex]
We combine the constant terms:
[tex]\[ \left(x + \frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 - \left(\frac{C}{2}\right)^2 - \left(\frac{D}{2}\right)^2 + E = 0 \][/tex]
[tex]\[ \left(x + \frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 = \left(\frac{C}{2}\right)^2 + \left(\frac{D}{2}\right)^2 - E \][/tex]
This gives us the standard form:
[tex]\[ \left(x + \frac{C}{2}\right)^2 + \left(y + \frac{D}{2}\right)^2 = r^2 \][/tex]
where:
[tex]\[ h = -\frac{C}{2}, \quad k = -\frac{D}{2}, \quad \text{and} \quad r^2 = \left(\frac{C}{2}\right)^2 + \left(\frac{D}{2}\right)^2 - E \][/tex]
### Step 3: Effect of Decreasing the Radius
If the radius [tex]\(r\)[/tex] is decreased, then [tex]\(r^2\)[/tex] becomes smaller. Given the equation:
[tex]\[ r^2 = \left(\frac{C}{2}\right)^2 + \left(\frac{D}{2}\right)^2 - E \][/tex]
we see that the terms [tex]\(\left(\frac{C}{2}\right)^2\)[/tex] and [tex]\(\left(\frac{D}{2}\right)^2\)[/tex] depend on [tex]\(C\)[/tex] and [tex]\(D\)[/tex], which are functions of the center coordinates. These coordinates are not changing, so [tex]\(C\)[/tex] and [tex]\(D\)[/tex] remain unchanged.
To decrease [tex]\(r^2\)[/tex], the term [tex]\(- E\)[/tex] must become larger (i.e., [tex]\(E\)[/tex] must increase since [tex]\( E\)[/tex] is subtracted).
### Conclusion
Therefore, the coefficients [tex]\(C\)[/tex] and [tex]\(D\)[/tex] are unchanged, but the coefficient [tex]\(E\)[/tex] increases when the radius is decreased.
The correct answer is:
[tex]\[ \boxed{E} \text{: $C$ and $D$ are unchanged, but $E$ increases.} \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
