Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Ask your questions and receive precise answers from experienced professionals across different disciplines. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
