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Does the point (3, -1) lie on the circle [tex]\((x+1)^2 + (y-1)^2 = 16\)[/tex]?

A. Yes; the point is represented by [tex]\((h, k)\)[/tex] in the equation.
B. Yes; when you plug the point in for [tex]\(x\)[/tex] and [tex]\(y\)[/tex], you get a true statement.
C. No; when you plug in the point for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the equation, you do not get a true statement.
D. No; the point is not represented by [tex]\((h, k)\)[/tex] in the equation.



Answer :

GinnyAnswer
To determine if the point [tex]\((3, -1)\)[/tex] lies on the circle given by the equation [tex]\( (x+1)^2 + (y-1)^2 = 16 \)[/tex], we substitute the coordinates of the point into the equation and check if the resulting statement is true. Here's the step-by-step process:

1. Circle's Equation:
[tex]\[(x+1)^2 + (y-1)^2 = 16\][/tex]

2. Substitute [tex]\(x = 3\)[/tex] and [tex]\(y = -1\)[/tex] into the equation:
[tex]\[ (3 + 1)^2 + (-1 - 1)^2 \][/tex]

3. Simplify the expression inside the parentheses first:
[tex]\[ (4)^2 + (-2)^2 \][/tex]

4. Compute the squares:
[tex]\[ 16 + 4 \][/tex]

5. Add the results:
[tex]\[ 20 \][/tex]

6. Compare the result with the right-hand side of the equation [tex]\( (x+1)^2 + (y-1)^2 = 16 \)[/tex]:
[tex]\[ 20 \neq 16 \][/tex]

Since the left-hand side of the equation [tex]\( 20 \)[/tex] does not equal the right-hand side [tex]\( 16 \)[/tex], the point [tex]\((3, -1)\)[/tex] does not satisfy the circle's equation.

Conclusion: The point [tex]\((3, -1)\)[/tex] does not lie on the circle. Therefore, the correct option is:

No; when you plug in the point for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the equation, you do not get a true statement.

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