Select the correct answer.

Which point lies on the circle represented by the equation [tex](x-3)^2+(y+4)^2=6^2[/tex]?

A. [tex](9, -2)[/tex]
B. [tex](0, 11)[/tex]
C. [tex](3, 10)[/tex]
D. [tex](-9, 4)[/tex]
E. [tex](-3, -4)[/tex]



Answer :

To determine which point lies on the circle represented by the equation [tex]\((x - 3)^2 + (y + 4)^2 = 6^2\)[/tex], we need to substitute the coordinates of each point into the equation of the circle and check if the equation holds true.

Let's start by recalling the general form of the equation of a circle:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius. For the given equation:
[tex]\[ (x - 3)^2 + (y + 4)^2 = 6^2 \][/tex]
the center of the circle is [tex]\((3, -4)\)[/tex] and the radius is [tex]\(6\)[/tex].

Now let's check each point to see if it satisfies the given equation.

### Option A: [tex]\((9, -2)\)[/tex]
Substitute [tex]\(x = 9\)[/tex] and [tex]\(y = -2\)[/tex]:
[tex]\[ (9 - 3)^2 + (-2 + 4)^2 = 6^2 \\ 6^2 + 2^2 = 36 \\ 36 + 4 = 40 \\ 40 \neq 36 \][/tex]
So, [tex]\((9, -2)\)[/tex] does not lie on the circle.

### Option B: [tex]\((0, 11)\)[/tex]
Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 11\)[/tex]:
[tex]\[ (0 - 3)^2 + (11 + 4)^2 = 6^2 \\ (-3)^2 + 15^2 = 36 \\ 9 + 225 = 234 \\ 234 \neq 36 \][/tex]
So, [tex]\((0, 11)\)[/tex] does not lie on the circle.

### Option C: [tex]\((3, 10)\)[/tex]
Substitute [tex]\(x = 3\)[/tex] and [tex]\(y = 10\)[/tex]:
[tex]\[ (3 - 3)^2 + (10 + 4)^2 = 6^2 \\ 0^2 + 14^2 = 36 \\ 0 + 196 = 196 \\ 196 \neq 36 \][/tex]
So, [tex]\((3, 10)\)[/tex] does not lie on the circle.

### Option D: [tex]\((-9, 4)\)[/tex]
Substitute [tex]\(x = -9\)[/tex] and [tex]\(y = 4\)[/tex]:
[tex]\[ (-9 - 3)^2 + (4 + 4)^2 = 6^2 \\ (-12)^2 + 8^2 = 36 \\ 144 + 64 = 208 \\ 208 \neq 36 \][/tex]
So, [tex]\((-9, 4)\)[/tex] does not lie on the circle.

### Option E: [tex]\((-3, -4)\)[/tex]
Substitute [tex]\(x = -3\)[/tex] and [tex]\(y = -4\)[/tex]:
[tex]\[ (-3 - 3)^2 + (-4 + 4)^2 = 6^2 \\ (-6)^2 + 0^2 = 36 \\ 36 + 0 = 36 \\ 36 = 36 \][/tex]
So, [tex]\((-3, -4)\)[/tex] does lie on the circle.

Based on this detailed examination, the point that lies on the circle is:
[tex]\[ \text{E. } (-3, -4) \][/tex]