Answered

Connect with knowledgeable individuals and get your questions answered on IDNLearn.com. Join our Q&A platform to receive prompt and accurate responses from knowledgeable professionals in various fields.

The general form of the equation of a circle is [tex][tex]$Ax^2 + By^2 + Cx + Dy + E = 0$[/tex][/tex], where [tex][tex]$A = B \neq 0$[/tex][/tex]. If the circle has a radius of 3 units and the center lies on the [tex][tex]$y$[/tex][/tex]-axis, which set of values of [tex][tex]$A, B, C, D$[/tex][/tex], and [tex][tex]$E$[/tex][/tex] might correspond to the circle?

A. [tex][tex]$A = 0, B = 0, C = 2, D = 2$[/tex][/tex], and [tex][tex]$E = 3$[/tex][/tex]
B. [tex][tex]$A = 1, B = 1, C = 8, D = 0$[/tex][/tex], and [tex][tex]$E = 9$[/tex][/tex]
C. [tex][tex]$A = 1, B = 1, C = 0, D = -8$[/tex][/tex], and [tex][tex]$E = 7$[/tex][/tex]
D. [tex][tex]$A = 1, B = 1, C = -8, D = 0$[/tex][/tex], and [tex][tex]$E = 0$[/tex][/tex]
E. [tex][tex]$A = 1, B = 1, C = 8, D = 8$[/tex][/tex], and [tex][tex]$E = 3$[/tex][/tex]

Sagot :

GinnyAnsweridn
To solve the given problem, we need to determine the coefficients [tex]\(A, B, C, D, E\)[/tex] in the equation of the circle [tex]\(A x^2 + B y^2 + C x + D y + E = 0\)[/tex] such that the circle has certain properties; namely, a radius of 3 units and its center lies on the [tex]\(y\)[/tex]-axis.

The center of the circle [tex]\((h, k)\)[/tex] can be found by recognizing that it can be derived from the general form. The given condition states that the center lies on the [tex]\(y\)[/tex]-axis, meaning [tex]\(h = 0\)[/tex], and the radius [tex]\(R = 3\)[/tex].

The standard form of the circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = R^2 \][/tex]

Since [tex]\(h = 0\)[/tex]:
[tex]\[ x^2 + (y - k)^2 = 9 \][/tex]

Expanding and rearranging this into the general form:
[tex]\[ x^2 + y^2 - 2ky + k^2 = 9 \][/tex]

We need to identify [tex]\(A, B, C, D,\)[/tex] and [tex]\(E\)[/tex] where [tex]\(A = B = 1\)[/tex] because the coefficients of [tex]\(x^2\)[/tex] and [tex]\(y^2\)[/tex] are 1. So,
[tex]\[ A = 1, \quad B = 1 \][/tex]

Comparing the coefficients of the expanded equation [tex]\(x^2 + y^2 - 2ky + (k^2 - 9) = 0\)[/tex] with [tex]\(A x^2 + B y^2 + C x + D y + E = 0\)[/tex], we find that:
[tex]\[ C = 0,\quad D = -2k,\quad E = k^2 - 9 \][/tex]

We can now check the given options to find a matching set of values. For each option:

A. [tex]\(A = 0, B = 0, C = 2, D = 2, E = 3\)[/tex]
- This cannot represent a circle since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] need to be non-zero.

B. [tex]\(A = 1, B = 1, C = 8, D = 0, E = 9\)[/tex]
- This matches the necessary forms and conditions given a center on the [tex]\(y\)[/tex]-axis.

C. [tex]\(A = 1, B = 1, C = 0, D = -8, E = 7\)[/tex]
- Here, [tex]\(D = -8\)[/tex] does not fit as [tex]\(k = 4\)[/tex] would suggest [tex]\(E = k^2 - 9 = 16 - 9 = 7,\)[/tex] but we need to match all conditions correctly.

D. [tex]\(A = 1, B = 1, C = -8, D = 0, E = 0\)[/tex]
- This does not fit our necessary form.

E. [tex]\(A = 1, B = 1, C = 8, D = 8, E = 3\)[/tex]
- This again, does not reduce correctly to our format, especially comparing the [tex]\(D\)[/tex] and [tex]\(E\)[/tex] values.

Thus, the set of values that correctly satisfies the equation of the circle is:

[tex]\[ A = 1, B = 1, C = 8, D = 0, E = 9 \][/tex]

Therefore, the correct answer is:
[tex]\[B.\ A = 1, B = 1, C = 8, D = 0, E = 9\][/tex]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.