IDNLearn.com: Your one-stop platform for getting reliable answers to any question. Join our knowledgeable community and access a wealth of reliable answers to your most pressing questions.
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.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
