Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Let's analyze the given situation step-by-step.
### Original Circle Equation
The original equation of the circle is:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
### Moving the Circle Horizontally to the Left
When we move a circle horizontally to the left by a distance [tex]\(a\)[/tex], only the x-coordinate of the center of the circle changes, which impacts the [tex]\(C\)[/tex] coefficient.
### New Circle Equation After Movement
Let's apply the horizontal shift to the circle's equation. Let the circle be shifted to the left by a distance [tex]\(a\)[/tex]. The new position of any point [tex]\((x, y)\)[/tex] on the circle will be replaced by [tex]\((x + a, y)\)[/tex].
### Substitute [tex]\((x + a)\)[/tex] for [tex]\(x\)[/tex]
We'll substitute [tex]\(x + a\)[/tex] for [tex]\(x\)[/tex] in the original equation. The new equation is:
[tex]\[ (x + a)^2 + y^2 + C(x + a) + Dy + E = 0. \][/tex]
### Expand and Simplify the New Equation:
Expanding this equation, we get:
[tex]\[ (x + a)^2 = x^2 + 2ax + a^2. \][/tex]
So,
[tex]\[ x^2 + 2ax + a^2 + y^2 + C(x + a) + Dy + E = 0. \][/tex]
[tex]\[ x^2 + 2ax + a^2 + y^2 + Cx + Ca + Dy + E = 0. \][/tex]
### Group the Terms:
Now, let's group the terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 + y^2 + (C+2a)x + Dy + (a^2 + Ca + E) = 0. \][/tex]
### Compare with Original Equation:
Let's compare the new equation:
[tex]\[ x^2 + y^2 + (C + 2a)x + Dy + (a^2 + Ca + E) = 0 \][/tex]
with the original equation:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
### Effect on Coefficients [tex]\(C\)[/tex] and [tex]\(D\)[/tex]:
1. Coefficient [tex]\(C\)[/tex]:
- The new coefficient of [tex]\(x\)[/tex] is [tex]\(C + 2a\)[/tex].
- Thus, [tex]\(C\)[/tex] changes to [tex]\(C + 2a\)[/tex].
2. Coefficient [tex]\(D\)[/tex]:
- The coefficient of [tex]\(y\)[/tex] remains [tex]\(D\)[/tex], as moving horizontally doesn't affect the [tex]\(y\)[/tex]-value.
In summary, when the circle is moved horizontally to the left by a distance [tex]\(a\)[/tex]:
- The coefficient [tex]\(C\)[/tex] increases by twice the distance shifted, i.e., it becomes [tex]\(C + 2a\)[/tex].
- The coefficient [tex]\(D\)[/tex] remains unchanged.
Therefore, the answer is:
When the circle is moved horizontally to the left, the coefficient [tex]\(C\)[/tex] increases by twice the distance shifted, and the coefficient [tex]\(D\)[/tex] remains unchanged.
### Original Circle Equation
The original equation of the circle is:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
### Moving the Circle Horizontally to the Left
When we move a circle horizontally to the left by a distance [tex]\(a\)[/tex], only the x-coordinate of the center of the circle changes, which impacts the [tex]\(C\)[/tex] coefficient.
### New Circle Equation After Movement
Let's apply the horizontal shift to the circle's equation. Let the circle be shifted to the left by a distance [tex]\(a\)[/tex]. The new position of any point [tex]\((x, y)\)[/tex] on the circle will be replaced by [tex]\((x + a, y)\)[/tex].
### Substitute [tex]\((x + a)\)[/tex] for [tex]\(x\)[/tex]
We'll substitute [tex]\(x + a\)[/tex] for [tex]\(x\)[/tex] in the original equation. The new equation is:
[tex]\[ (x + a)^2 + y^2 + C(x + a) + Dy + E = 0. \][/tex]
### Expand and Simplify the New Equation:
Expanding this equation, we get:
[tex]\[ (x + a)^2 = x^2 + 2ax + a^2. \][/tex]
So,
[tex]\[ x^2 + 2ax + a^2 + y^2 + C(x + a) + Dy + E = 0. \][/tex]
[tex]\[ x^2 + 2ax + a^2 + y^2 + Cx + Ca + Dy + E = 0. \][/tex]
### Group the Terms:
Now, let's group the terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 + y^2 + (C+2a)x + Dy + (a^2 + Ca + E) = 0. \][/tex]
### Compare with Original Equation:
Let's compare the new equation:
[tex]\[ x^2 + y^2 + (C + 2a)x + Dy + (a^2 + Ca + E) = 0 \][/tex]
with the original equation:
[tex]\[ x^2 + y^2 + Cx + Dy + E = 0. \][/tex]
### Effect on Coefficients [tex]\(C\)[/tex] and [tex]\(D\)[/tex]:
1. Coefficient [tex]\(C\)[/tex]:
- The new coefficient of [tex]\(x\)[/tex] is [tex]\(C + 2a\)[/tex].
- Thus, [tex]\(C\)[/tex] changes to [tex]\(C + 2a\)[/tex].
2. Coefficient [tex]\(D\)[/tex]:
- The coefficient of [tex]\(y\)[/tex] remains [tex]\(D\)[/tex], as moving horizontally doesn't affect the [tex]\(y\)[/tex]-value.
In summary, when the circle is moved horizontally to the left by a distance [tex]\(a\)[/tex]:
- The coefficient [tex]\(C\)[/tex] increases by twice the distance shifted, i.e., it becomes [tex]\(C + 2a\)[/tex].
- The coefficient [tex]\(D\)[/tex] remains unchanged.
Therefore, the answer is:
When the circle is moved horizontally to the left, the coefficient [tex]\(C\)[/tex] increases by twice the distance shifted, and the coefficient [tex]\(D\)[/tex] remains unchanged.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.