Question

determine the distance between \\((3, 5)\\) and \\((-2, -3)\\).

what is the length of the radius of the circle?

which equation matches the graph?
a. \\((x - 4)^2 + (y + 2)^2 = 4\\)
b. \\((x - 4)^2 + (y + 2)^2 = 16\\)
c. \\((x + 4)^2 + (y - 2)^2 = 4\\)
d. \\((x + 4)^2 + (y - 2)^2 = 16\\)

problems 4-5: here is a circle with center \\((2, 0)\\).
is \\((3, \sqrt{15})\\) on the circle? circle one.
yes no
show or explain your thinking.

is \\((5, 2)\\) on the circle? circle one.
yes no
show or explain your thinking.

Explanation:

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<reasoning>

Calculate distance between points

Using the Distance Formula
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substitute \((3, 5)\) and \((-2, -3)\):
\[
d = \sqrt{(-2 - 3)^2 + (-3 - 5)^2} = \sqrt{(-5)^2 + (-8)^2} = \sqrt{25 + 64} = \sqrt{89}
\]

Find the radius of the circle

Using the Circle Radius and Diameter knowledge point
The center is at \((4, 2)\) and it passes through \((1, 0)\).
\[
r = \sqrt{(1 - 4)^2 + (0 - 2)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]

Determine the matching circle equation

The center is at \((-4, 2)\) and it passes through \((0, 2)\).
The radius is the distance from \((-4, 2)\) to \((0, 2)\):
\[
r = 0 - (-4) = 4
\]
The standard equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substitute \(h = -4\), \(k = 2\), and \(r = 4\):
\[
(x + 4)^2 + (y - 2)^2 = 16
\]
This matches option D.

Check if point is on the circle

The circle has center \((2, 0)\) and passes through \((6, 0)\).
The radius is:
\[
r = 6 - 2 = 4
\]
The equation of the circle is:
\[
(x - 2)^2 + y^2 = 16
\]
Substitute \((3, \sqrt{15})\):
\[
(3 - 2)^2 + (\sqrt{15})^2 = 1^2 + 15 = 16
\]
Since \(16 = 16\), the point is on the circle.

Check if second point is on the circle

Using the circle equation \((x - 2)^2 + y^2 = 16\), substitute \((5, 2)\):
\[
(5 - 2)^2 + 2^2 = 3^2 + 4 = 9 + 4 = 13
\]
Since \(13
eq 16\), the point is not on the circle.
</reasoning>

<answer>

Question 1

\(\sqrt{89}\)

Question 2

\(\sqrt{13}\)

Question 3

D. \((x + 4)^2 + (y - 2)^2 = 16\)

Question 4

Yes.
Since the center is \((2, 0)\) and it passes through \((6, 0)\), the radius is \(4\). The equation of the circle is \((x - 2)^2 + y^2 = 16\). Substituting \((3, \sqrt{15})\) gives \((3 - 2)^2 + (\sqrt{15})^2 = 1 + 15 = 16\), which is true.

Question 5

No.
Using the circle equation \((x - 2)^2 + y^2 = 16\), substituting \((5, 2)\) gives \((5 - 2)^2 + 2^2 = 9 + 4 = 13\), which does not equal \(16\).
</answer>

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Answer:

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<reasoning>

Calculate distance between points

Using the Distance Formula
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substitute \((3, 5)\) and \((-2, -3)\):
\[
d = \sqrt{(-2 - 3)^2 + (-3 - 5)^2} = \sqrt{(-5)^2 + (-8)^2} = \sqrt{25 + 64} = \sqrt{89}
\]

Find the radius of the circle

Using the Circle Radius and Diameter knowledge point
The center is at \((4, 2)\) and it passes through \((1, 0)\).
\[
r = \sqrt{(1 - 4)^2 + (0 - 2)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]

Determine the matching circle equation

The center is at \((-4, 2)\) and it passes through \((0, 2)\).
The radius is the distance from \((-4, 2)\) to \((0, 2)\):
\[
r = 0 - (-4) = 4
\]
The standard equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substitute \(h = -4\), \(k = 2\), and \(r = 4\):
\[
(x + 4)^2 + (y - 2)^2 = 16
\]
This matches option D.

Check if point is on the circle

The circle has center \((2, 0)\) and passes through \((6, 0)\).
The radius is:
\[
r = 6 - 2 = 4
\]
The equation of the circle is:
\[
(x - 2)^2 + y^2 = 16
\]
Substitute \((3, \sqrt{15})\):
\[
(3 - 2)^2 + (\sqrt{15})^2 = 1^2 + 15 = 16
\]
Since \(16 = 16\), the point is on the circle.

Check if second point is on the circle

Using the circle equation \((x - 2)^2 + y^2 = 16\), substitute \((5, 2)\):
\[
(5 - 2)^2 + 2^2 = 3^2 + 4 = 9 + 4 = 13
\]
Since \(13
eq 16\), the point is not on the circle.
</reasoning>

<answer>

Question 1

\(\sqrt{89}\)

Question 2

\(\sqrt{13}\)

Question 3

D. \((x + 4)^2 + (y - 2)^2 = 16\)

Question 4

Yes.
Since the center is \((2, 0)\) and it passes through \((6, 0)\), the radius is \(4\). The equation of the circle is \((x - 2)^2 + y^2 = 16\). Substituting \((3, \sqrt{15})\) gives \((3 - 2)^2 + (\sqrt{15})^2 = 1 + 15 = 16\), which is true.

Question 5

No.
Using the circle equation \((x - 2)^2 + y^2 = 16\), substituting \((5, 2)\) gives \((5 - 2)^2 + 2^2 = 9 + 4 = 13\), which does not equal \(16\).
</answer>

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