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

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.

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

Since $\dfrac{1 + B_x}{2} = 4$, we get $B_x = 8 - 1 = 7$.

Parallel lines have the same slope, so $m = \dfrac{1}{2}$. Substitute $(6, 1)$ into $y = \dfrac{1}{2}x + b$: $1 = \dfrac{1}{2}(6) + b = 3 + b$, so $b = -2$. The equation is $y = \dfrac{1}{2}x - 2\textsf{.}$

Perpendicular lines have slopes that are negative reciprocals. The slope of line $t$ is $5$, so the slope of line $k$ is $-\dfrac{1}{5}\textsf{.}$

The endpoints share the same $y$-coordinate, so the diameter is horizontal with length $|12 - (-6)| = 18\textsf{.}$ The radius is half the diameter, so $r = \dfrac{18}{2} = 9\textsf{.}$