Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

The equation of circle [tex]\( A \)[/tex] is [tex]\( x^2 + y^2 + Cx + Dy + E = 0 \)[/tex]. If the circle is moved horizontally to the left of the [tex]\( y \)[/tex]-axis without changing the radius, how are the coefficients [tex]\( C \)[/tex] and [tex]\( D \)[/tex] affected?

Sagot :

GinnyAnswer
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.
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.