NJSLA Algebra II Expressing Geometric Properties with Equations. Practice it free.

The diameter length is $\sqrt{(5 - (-5))^2 + (26 - 2)^2} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26\textsf{.}$ The radius is half the diameter, so $r = \dfrac{26}{2} = 13\textsf{.}$

Circle A has center $(-5, 3)$ and radius $5$. Shifting right $2$ moves the center to $(-3, 3)$, so Circle B is $(x + 3)^2 + (y - 3)^2 = 25$.

The center is $(2, 8)$ and the radius is $9\textsf{.}$ A point on the circle must satisfy $2 - 9 \le a \le 2 + 9\textsf{,}$ so $-7 \le a \le 11\textsf{.}$ The value $5$ is in this interval.

The right side is $r^{2} = 25\textsf{,}$ so the radius measures $5$ units.

$y$-coordinate $= \dfrac{10 + 2}{2} = \dfrac{12}{2} = 6$.

Perpendicular lines have slopes that are negative reciprocals, so $m = \dfrac{3}{2}$. Substitute $(6, -1)$ into $y = \dfrac{3}{2}x + b$: $-1 = \dfrac{3}{2}(6) + b = 9 + b$, so $b = -10$. The equation is $y = \dfrac{3}{2}x - 10\textsf{.}$

Parallel lines have the same slope. In $y = -5x + 2$, the slope is $-5$, so the slope of line $q$ is also $-5\textsf{.}$

Vertex form $y = (x - h)^{2} + k$ has vertex $(h, k)$, so the vertex is $(3, 5)$ and the x-coordinate is $3$.