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Write the standard form of the equation of the circle with the given characteristics:

Endpoints of a diameter: [tex](2,3),(-12,-13)[/tex]

[tex]\square[/tex]


Sagot :

To find the standard form of the equation of a circle given the endpoints of a diameter, we need to determine the center and the radius of the circle.

Step 1: Calculate the center of the circle

The center of the circle, [tex]\( (h, k) \)[/tex], is the midpoint of the diameter's endpoints [tex]\((2, 3)\)[/tex] and [tex]\((-12, -13)\)[/tex].

The midpoint formula is:
[tex]\[ (h, k) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]

Substituting the given points:
[tex]\[ h = \frac{2 + (-12)}{2} = \frac{2 - 12}{2} = \frac{-10}{2} = -5 \][/tex]
[tex]\[ k = \frac{3 + (-13)}{2} = \frac{3 - 13}{2} = \frac{-10}{2} = -5 \][/tex]

So, the center of the circle is [tex]\((-5, -5)\)[/tex].

Step 2: Calculate the radius of the circle

The radius is half the length of the diameter. First, we find the length of the diameter using the distance formula between the endpoints [tex]\((2, 3)\)[/tex] and [tex]\((-12, -13)\)[/tex]:

The distance formula is:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Substituting the given points:
[tex]\[ \text{Distance} = \sqrt{(-12 - 2)^2 + (-13 - 3)^2} = \sqrt{(-14)^2 + (-16)^2} = \sqrt{196 + 256} = \sqrt{452} \][/tex]

Thus, the radius [tex]\( r \)[/tex] is half of the diameter:
[tex]\[ r = \frac{\sqrt{452}}{2} = \frac{21.2602916254693}{2} \approx 10.63014581273465 \][/tex]

Step 3: Write the standard form of 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]

Substitute [tex]\(h = -5\)[/tex], [tex]\(k = -5\)[/tex], and [tex]\(r \approx 10.63014581273465\)[/tex]:
[tex]\[ (x + 5)^2 + (y + 5)^2 = (10.63014581273465)^2 \][/tex]

Squaring the radius:
[tex]\[ (10.63014581273465)^2 \approx 113.00000000000001 \][/tex]

So, the standard form of the equation of the circle is:
[tex]\[ (x + 5)^2 + (y + 5)^2 = 113 \][/tex]