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Sagot :
The equation of a circle in its general form is given by:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0 \][/tex]
To understand how a horizontal move to the left affects the coefficients [tex]\(C\)[/tex] and [tex]\(D\)[/tex], we first need to convert this equation to the center-radius form and then see how the shift modifies it. Let's proceed step-by-step:
### Step 1: Convert to Center-Radius Form
The general form [tex]\( x^2 + y^2 + Cx + Dy + E = 0 \)[/tex] can be completed to match the form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + Cx \to (x + \frac{C}{2})^2 - \left( \frac{C}{2} \right)^2 \][/tex]
[tex]\[ y^2 + Dy \to (y + \frac{D}{2})^2 - \left( \frac{D}{2} \right)^2 \][/tex]
So, the equation becomes:
[tex]\[ (x + \frac{C}{2})^2 - \left( \frac{C}{2} \right)^2 + (y + \frac{D}{2})^2 - \left( \frac{D}{2} \right)^2 + E = 0 \][/tex]
Combine the constant terms to isolate the squared form:
[tex]\[ (x + \frac{C}{2})^2 + (y + \frac{D}{2})^2 = \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E \][/tex]
This gives us the center of the circle [tex]\((- \frac{C}{2}, - \frac{D}{2})\)[/tex] and radius [tex]\(r = \sqrt{\left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E}\)[/tex].
### Step 2: Translate the Circle
If the circle is moved horizontally to the left by [tex]\(a\)[/tex] units, the new center will be:
[tex]\[ \left(- \frac{C}{2} - a, - \frac{D}{2} \right) \][/tex]
### Step 3: New Equation of Circle
With the new center, the equation of the circle in center-radius form is:
[tex]\[ \left( x - \left( -\frac{C}{2} - a \right) \right)^2 + \left( y + \frac{D}{2} \right)^2 = r^2 \][/tex]
[tex]\[ \left( x + \frac{C}{2} + a \right)^2 + \left( y + \frac{D}{2} \right)^2 = r^2 \][/tex]
### Step 4: Convert Back to General Form
Expand and simplify this new equation:
[tex]\[ \left( x + \frac{C}{2} + a \right)^2 + \left( y + \frac{D}{2} \right)^2 = r^2 \][/tex]
[tex]\[ x^2 + 2x\left( \frac{C}{2} + a \right) + \left( \frac{C}{2} + a \right)^2 + y^2 + Dy + \left( \frac{D}{2} \right)^2 = r^2\][/tex]
Comparing this with the standard form [tex]\( x^2 + y^2 + C'x + D'y + E' = 0 \)[/tex]:
[tex]\[ x^2 + y^2 + 2x\left( \frac{C}{2} + a \right) + C + k\][/tex]
We can see that the new coefficients [tex]\(C'\)[/tex] and [tex]\(D'\)[/tex] are given by:
[tex]\[ C' = 2\left( \frac{C}{2} + a \right) = C + 2a \][/tex]
The coefficient [tex]\( D \)[/tex] does not change, as there was no vertical shift:
[tex]\[ D' = D \][/tex]
### Conclusion
By translating the circle horizontally to the left by [tex]\(a\)[/tex] units:
- The new coefficient [tex]\( C' \)[/tex] becomes [tex]\( C + 2a \)[/tex]
- The coefficient [tex]\( D \)[/tex] remains unchanged.
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0 \][/tex]
To understand how a horizontal move to the left affects the coefficients [tex]\(C\)[/tex] and [tex]\(D\)[/tex], we first need to convert this equation to the center-radius form and then see how the shift modifies it. Let's proceed step-by-step:
### Step 1: Convert to Center-Radius Form
The general form [tex]\( x^2 + y^2 + Cx + Dy + E = 0 \)[/tex] can be completed to match the form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + Cx \to (x + \frac{C}{2})^2 - \left( \frac{C}{2} \right)^2 \][/tex]
[tex]\[ y^2 + Dy \to (y + \frac{D}{2})^2 - \left( \frac{D}{2} \right)^2 \][/tex]
So, the equation becomes:
[tex]\[ (x + \frac{C}{2})^2 - \left( \frac{C}{2} \right)^2 + (y + \frac{D}{2})^2 - \left( \frac{D}{2} \right)^2 + E = 0 \][/tex]
Combine the constant terms to isolate the squared form:
[tex]\[ (x + \frac{C}{2})^2 + (y + \frac{D}{2})^2 = \left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E \][/tex]
This gives us the center of the circle [tex]\((- \frac{C}{2}, - \frac{D}{2})\)[/tex] and radius [tex]\(r = \sqrt{\left( \frac{C}{2} \right)^2 + \left( \frac{D}{2} \right)^2 - E}\)[/tex].
### Step 2: Translate the Circle
If the circle is moved horizontally to the left by [tex]\(a\)[/tex] units, the new center will be:
[tex]\[ \left(- \frac{C}{2} - a, - \frac{D}{2} \right) \][/tex]
### Step 3: New Equation of Circle
With the new center, the equation of the circle in center-radius form is:
[tex]\[ \left( x - \left( -\frac{C}{2} - a \right) \right)^2 + \left( y + \frac{D}{2} \right)^2 = r^2 \][/tex]
[tex]\[ \left( x + \frac{C}{2} + a \right)^2 + \left( y + \frac{D}{2} \right)^2 = r^2 \][/tex]
### Step 4: Convert Back to General Form
Expand and simplify this new equation:
[tex]\[ \left( x + \frac{C}{2} + a \right)^2 + \left( y + \frac{D}{2} \right)^2 = r^2 \][/tex]
[tex]\[ x^2 + 2x\left( \frac{C}{2} + a \right) + \left( \frac{C}{2} + a \right)^2 + y^2 + Dy + \left( \frac{D}{2} \right)^2 = r^2\][/tex]
Comparing this with the standard form [tex]\( x^2 + y^2 + C'x + D'y + E' = 0 \)[/tex]:
[tex]\[ x^2 + y^2 + 2x\left( \frac{C}{2} + a \right) + C + k\][/tex]
We can see that the new coefficients [tex]\(C'\)[/tex] and [tex]\(D'\)[/tex] are given by:
[tex]\[ C' = 2\left( \frac{C}{2} + a \right) = C + 2a \][/tex]
The coefficient [tex]\( D \)[/tex] does not change, as there was no vertical shift:
[tex]\[ D' = D \][/tex]
### Conclusion
By translating the circle horizontally to the left by [tex]\(a\)[/tex] units:
- The new coefficient [tex]\( C' \)[/tex] becomes [tex]\( C + 2a \)[/tex]
- The coefficient [tex]\( D \)[/tex] remains unchanged.
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