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Sagot :
To determine the equation of the circle in standard form, we need to find the center and the radius of the circle using the given diameter endpoints [tex]\( P(-10, 10) \)[/tex] and [tex]\( Q(6, -2) \)[/tex].
### Step-by-Step Solution
#### 1. Find the midpoint (center) of the circle:
The midpoint formula for points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
For points [tex]\( P(-10, 10) \)[/tex] and [tex]\( Q(6, -2) \)[/tex]:
- Midpoint [tex]\( x \)[/tex]-coordinate is:
[tex]\[ \frac{-10 + 6}{2} = \frac{-4}{2} = -2 \][/tex]
- Midpoint [tex]\( y \)[/tex]-coordinate is:
[tex]\[ \frac{10 - 2}{2} = \frac{8}{2} = 4 \][/tex]
So, the center of the circle is [tex]\( (-2, 4) \)[/tex].
#### 2. Find the radius of the circle:
The radius is half the distance between points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]. First, we calculate the distance between these points using the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting [tex]\( P(-10, 10) \)[/tex] and [tex]\( Q(6, -2) \)[/tex]:
[tex]\[ d = \sqrt{(6 - (-10))^2 + (-2 - 10)^2} \][/tex]
[tex]\[ d = \sqrt{(6 + 10)^2 + (-2 - 10)^2} \][/tex]
[tex]\[ d = \sqrt{16^2 + (-12)^2} \][/tex]
[tex]\[ d = \sqrt{256 + 144} \][/tex]
[tex]\[ d = \sqrt{400} \][/tex]
[tex]\[ d = 20 \][/tex]
The radius [tex]\( r \)[/tex] is half this distance:
[tex]\[ r = \frac{20}{2} = 10 \][/tex]
#### 3. Write the equation of the circle:
The standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
Substituting the center [tex]\((-2, 4)\)[/tex] and radius [tex]\(10\)[/tex] into the equation:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 10^2 \][/tex]
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
#### 4. Match with the given options:
Comparing this equation with the given options, we see that it matches option B:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
Thus, the correct answer is:
B. [tex]\((x+2)^2+(y-4)^2=100\)[/tex]
### Step-by-Step Solution
#### 1. Find the midpoint (center) of the circle:
The midpoint formula for points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
For points [tex]\( P(-10, 10) \)[/tex] and [tex]\( Q(6, -2) \)[/tex]:
- Midpoint [tex]\( x \)[/tex]-coordinate is:
[tex]\[ \frac{-10 + 6}{2} = \frac{-4}{2} = -2 \][/tex]
- Midpoint [tex]\( y \)[/tex]-coordinate is:
[tex]\[ \frac{10 - 2}{2} = \frac{8}{2} = 4 \][/tex]
So, the center of the circle is [tex]\( (-2, 4) \)[/tex].
#### 2. Find the radius of the circle:
The radius is half the distance between points [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]. First, we calculate the distance between these points using the distance formula:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting [tex]\( P(-10, 10) \)[/tex] and [tex]\( Q(6, -2) \)[/tex]:
[tex]\[ d = \sqrt{(6 - (-10))^2 + (-2 - 10)^2} \][/tex]
[tex]\[ d = \sqrt{(6 + 10)^2 + (-2 - 10)^2} \][/tex]
[tex]\[ d = \sqrt{16^2 + (-12)^2} \][/tex]
[tex]\[ d = \sqrt{256 + 144} \][/tex]
[tex]\[ d = \sqrt{400} \][/tex]
[tex]\[ d = 20 \][/tex]
The radius [tex]\( r \)[/tex] is half this distance:
[tex]\[ r = \frac{20}{2} = 10 \][/tex]
#### 3. Write the equation of the circle:
The standard form of the equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
Substituting the center [tex]\((-2, 4)\)[/tex] and radius [tex]\(10\)[/tex] into the equation:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 10^2 \][/tex]
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
#### 4. Match with the given options:
Comparing this equation with the given options, we see that it matches option B:
[tex]\[ (x + 2)^2 + (y - 4)^2 = 100 \][/tex]
Thus, the correct answer is:
B. [tex]\((x+2)^2+(y-4)^2=100\)[/tex]
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