PARCC Geometry Expressing Geometric Properties with Equations. Practice it free.

Sector area $= \dfrac{120}{360} \times \pi r^2 = \dfrac{120}{360} \times 3 \times 36 = 36$.

The area of the whole circle is $\pi r^{2} = \pi \cdot 10^{2} = 100\pi\textsf{.}$ A $90^{\circ}$ sector is $\dfrac{90}{360} = \dfrac{1}{4}$ of the circle, so the area is $\dfrac{1}{4} \cdot 100\pi = 25\pi\text{ mm}^{2}\textsf{.}$

The endpoints share the same $x$-coordinate, so the diameter is vertical with length $|6 - 0| = 6\textsf{.}$ The radius is half the diameter, so $r = \dfrac{6}{2} = 3\textsf{.}$

Circle A has center $(0, -2)$ and radius $4$. Shifting down $5$ moves the center to $(0, -7)$, so Circle B is $x^2 + (y + 7)^2 = 16$.

The center is $(-6, 4)$ and the radius is $10\textsf{.}$ A point on the circle must satisfy $-6 - 10 \le a \le -6 + 10\textsf{,}$ so $-16 \le a \le 4\textsf{.}$ The value $-11$ is in this interval.

Set $y = 0$: then $x^{2} = 25$, and the positive solution is $x = 5$.

In standard form the right side equals $r^{2}\textsf{,}$ so $r^{2} = 81\textsf{.}$

Let $B = (x, y)\textsf{.}$ Since $M = (4, 7)$ is the midpoint of $A = (1, 5)$ and $B\textsf{,}$ the midpoint formula gives $\dfrac{1 + x}{2} = 4$ and $\dfrac{5 + y}{2} = 7\textsf{.}$ Multiply $\dfrac{1 + x}{2} = 4$ by $2$ to get $1 + x = 8\textsf{,}$ so $x = 7\textsf{.}$ Multiply $\dfrac{5 + y}{2} = 7$ by $2$ to get $5 + y = 14\textsf{,}$ so $y = 9\textsf{.}$ The coordinates of $B$ are $(7, 9)\textsf{.}$