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Sagot :
To determine how the coefficients [tex]\( C \)[/tex], [tex]\( D \)[/tex], and [tex]\( E \)[/tex] are affected when the radius of the circle is decreased without changing the center coordinates, let's start by understanding the general equation of a circle.
The given equation of a circle is:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
First, we need to complete the square to find the standard form of the circle's equation, [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.
Rewrite the equation:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
To complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 + Cx + y^2 + Dy = -E. \][/tex]
[tex]\[ (x^2 + Cx) + (y^2 + Dy) = -E. \][/tex]
Completing the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 + Cx = (x + \frac{C}{2})^2 - \left( \frac{C}{2} \right)^2. \][/tex]
Completing the square for [tex]\(y\)[/tex]:
[tex]\[ y^2 + Dy = (y + \frac{D}{2})^2 - \left( \frac{D}{2} \right)^2. \][/tex]
Substitute these back into the original equation:
[tex]\[ \left( x + \frac{C}{2} \right)^2 - \left( \frac{C}{2} \right)^2 + \left( y + \frac{D}{2} \right)^2 - \left( \frac{D}{2} \right)^2 = -E. \][/tex]
[tex]\[ \left( x + \frac{C}{2} \right)^2 + \left( y + \frac{D}{2} \right)^2 = -E + \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2. \][/tex]
Let’s denote the center of the circle as:
[tex]\[ h = -\frac{C}{2}, \][/tex]
[tex]\[ k = -\frac{D}{2}, \][/tex]
And let the radius [tex]\( r \)[/tex] be:
[tex]\[ r^2 = \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E. \][/tex]
If we decrease the radius [tex]\( r \)[/tex], the term on the right-hand side of the equation [tex]\(\left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E\)[/tex] also decreases.
Let’s denote the new term after decreasing the radius by [tex]\( E' \)[/tex]:
[tex]\[ r'^2 = \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E'. \][/tex]
Given that [tex]\( r'^2 < r^2 \)[/tex], we have:
[tex]\[ E' > E. \][/tex]
Therefore, the only change involves [tex]\( E \)[/tex], which will increase. The coefficients [tex]\( C \)[/tex] and [tex]\( D \)[/tex] remain the same because the center [tex]\((h, k)\)[/tex] of the circle does not change.
So, the answer is:
E. [tex]\( C \)[/tex] and [tex]\( D \)[/tex] are unchanged, but [tex]\( E \)[/tex] increases.
The given equation of a circle is:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
First, we need to complete the square to find the standard form of the circle's equation, [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.
Rewrite the equation:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
To complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 + Cx + y^2 + Dy = -E. \][/tex]
[tex]\[ (x^2 + Cx) + (y^2 + Dy) = -E. \][/tex]
Completing the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 + Cx = (x + \frac{C}{2})^2 - \left( \frac{C}{2} \right)^2. \][/tex]
Completing the square for [tex]\(y\)[/tex]:
[tex]\[ y^2 + Dy = (y + \frac{D}{2})^2 - \left( \frac{D}{2} \right)^2. \][/tex]
Substitute these back into the original equation:
[tex]\[ \left( x + \frac{C}{2} \right)^2 - \left( \frac{C}{2} \right)^2 + \left( y + \frac{D}{2} \right)^2 - \left( \frac{D}{2} \right)^2 = -E. \][/tex]
[tex]\[ \left( x + \frac{C}{2} \right)^2 + \left( y + \frac{D}{2} \right)^2 = -E + \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2. \][/tex]
Let’s denote the center of the circle as:
[tex]\[ h = -\frac{C}{2}, \][/tex]
[tex]\[ k = -\frac{D}{2}, \][/tex]
And let the radius [tex]\( r \)[/tex] be:
[tex]\[ r^2 = \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E. \][/tex]
If we decrease the radius [tex]\( r \)[/tex], the term on the right-hand side of the equation [tex]\(\left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E\)[/tex] also decreases.
Let’s denote the new term after decreasing the radius by [tex]\( E' \)[/tex]:
[tex]\[ r'^2 = \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E'. \][/tex]
Given that [tex]\( r'^2 < r^2 \)[/tex], we have:
[tex]\[ E' > E. \][/tex]
Therefore, the only change involves [tex]\( E \)[/tex], which will increase. The coefficients [tex]\( C \)[/tex] and [tex]\( D \)[/tex] remain the same because the center [tex]\((h, k)\)[/tex] of the circle does not change.
So, the answer is:
E. [tex]\( C \)[/tex] and [tex]\( D \)[/tex] are unchanged, but [tex]\( E \)[/tex] increases.
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