Question

Даны точки А (-3;0) и В (3;6).Написать уравнение окружности, диаметром которой служит отрезок АВподробно пожалуйст

Answer 1

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Answer 2

I can help you with your math question. Here is the detailed solution:

The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. To find the equation of a circle that has the segment $AB$ as its diameter, we need to find the coordinates of the center and the length of the radius.

The center of the circle is the midpoint of the segment $AB$. We can use the midpoint formula to find it:

$$ \begin{aligned} h &= \frac{x_A + x_B}{2} = \frac{-3 + 3}{2} = 0 \\ k &= \frac{y_A + y_B}{2} = \frac{0 + 6}{2} = 3 \end{aligned} $$

The radius of the circle is half of the length of the segment $AB$. We can use the distance formula to find it:

$$ \begin{aligned} r &= \frac{1}{2} \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \\ &= \frac{1}{2} \sqrt{(3 - (-3))^2 + (6 - 0)^2} \\ &= \frac{1}{2} \sqrt{36 + 36} \\ &= \frac{1}{2} \sqrt{72} \\ &= \frac{1}{2} \cdot 6 \sqrt{2} \\ &= 3 \sqrt{2} \end{aligned} $$

Now we can substitute the values of $h$, $k$, and $r$ into the equation of the circle:

$$ \begin{aligned} (x-h)^2 + (y-k)^2 &= r^2 \\ (x-0)^2 + (y-3)^2 &= (3 \sqrt{2})^2 \\ x^2 + (y-3)^2 &= 18 \end{aligned} $$

This is the equation of the circle that has the segment $AB$ as its diameter. I hope this helps.Ответь+подробно.+Даны+точки+А+(-3;0)+и+В+(3;6).Написать+уравнение+окружности,+диаметром+которой+служит+отрезок+АВподробно+пожалуйст