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QUESTION 15

If the equation of a circle in General Form is [tex]$x^2+y^2-2x+6y-6=0$[/tex], which equation of a circle in Standard Form does this represent?

A. [tex]$(x+1)^2+(y-3)^2=46$[/tex]
B. [tex][tex]$(x-1)^2+(y+3)^2=16$[/tex][/tex]
C. [tex]$(x-2)^2+(y+6)^2=46$[/tex]
D. [tex]$(x+2)^2+(y-6)^2=16$[/tex]
E. [tex][tex]$x^2+(y-3)^2=16$[/tex][/tex]

QUESTION 16

If the polygon pictured below is placed on the coordinate plane as seen, and is rotated about the [tex]$x$[/tex]-axis, which sketch is a representative of it after it has been rotated 360 degrees about the [tex]$x$[/tex]-axis?


Sagot :

Let's focus on question 15, which involves converting the equation of a circle from its General Form to its Standard Form.

### Step-by-Step Solution:

1. Equation in General Form:
The given equation of the circle in General Form is:
[tex]\[ x^2 + y^2 - 2x + 6y - 6 = 0 \][/tex]

2. Grouping the Terms:
Group the [tex]\(x\)[/tex] terms and [tex]\(y\)[/tex] terms together:
[tex]\[ (x^2 - 2x) + (y^2 + 6y) = 6 \][/tex]

3. Completing the Square on the [tex]\(x\)[/tex] Terms:
- Take the coefficient of [tex]\(x\)[/tex], which is [tex]\(-2\)[/tex], halve it to get [tex]\(-1\)[/tex], and then square it to get [tex]\(1\)[/tex].
- Add and subtract [tex]\(1\)[/tex] to complete the square:
[tex]\[ (x^2 - 2x + 1 - 1) = (x - 1)^2 - 1 \][/tex]

4. Completing the Square on the [tex]\(y\)[/tex] Terms:
- Take the coefficient of [tex]\(y\)[/tex], which is [tex]\(6\)[/tex], halve it to get 3, and then square it to get 9.
- Add and subtract 9 to complete the square:
[tex]\[ (y^2 + 6y + 9 - 9) = (y + 3)^2 - 9 \][/tex]

5. Rewriting the Equation with Completed Squares:
Substitute the completed square terms back into the equation:
[tex]\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 = 6 \][/tex]

6. Simplifying the Equation:
Combine constants on the right side to convert to the Standard Form:
[tex]\[ (x - 1)^2 + (y + 3)^2 - 10 = 6 \][/tex]
[tex]\[ (x - 1)^2 + (y + 3)^2 = 16 \][/tex]

7. Conclusion:
The equation in Standard Form is:
[tex]\[ (x - 1)^2 + (y + 3)^2 = 16 \][/tex]

### Final Answer:
Therefore, the equation of the circle in Standard Form is:
[tex]\[ B. (x - 1)^2 + (y + 3)^2 = 16 \][/tex]

So, the correct option is B.