overline{AB} is a diameter of odot(P, r). If | vedclass

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(A) The center $P$ of a circle is the midpoint of its diameter $\overline{AB}$.Given $A(x_1, y_1) = (4, 0)$ and $B(x_2, y_2) = (-2, 2)$.The midpoint f

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$(1, 1)$

Vedclass · Answer

(A) The center $P$ of a circle is the midpoint of its diameter $\overline{AB}$.Given $A(x_1, y_1) = (4, 0)$ and $B(x_2, y_2) = (-2, 2)$.The midpoint formula is $P(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.Substituting the values: $P(x, y) = \left( \frac{4 + (-2)}{2}, \frac{0 + 2}{2} \right)$.$P(x, y) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)$.Therefore,the coordinates of $P$ are $(1, 1)$.

$\overline{AB}$ is a diameter of $\odot(P, r)$. If $A(4, 0)$ and $B(-2, 2)$ are given,then the coordinates of $P$ are.............

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