Answer :
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]
