(By the property right-angle in a semicircle, $AC$ is the diameter of the circle)

\begin{align*} \text{Length of } AC & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ (7 - 1)^2 + (12 - 4)^2 } \\ & = 10 \text{ units} \\ \\ \text{Radius} & = {10 \over 2} = 5 \text{ units} \\ \\ \text{Midpoint of } AC & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {1 + 7 \over 2}, {4 + 12 \over 2} \right) \\ & = (4, 8) \\ \\ \text{Centre: } & (4, 8) \\ \\ \text{Eqn: } (x - 4)^2 & + (y - 8)^2 = 5^2 \\ (x - 4)^2 & + (y - 8)^2 = 25 \end{align*}

(The line and the perpendicular bisector will meet at the centre)

\begin{align*} \text{Midpoint of } (-1, 6) \text{ and } (2, 3) & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {-1 + 2 \over 2}, {6 + 3 \over 2} \right) \\ & = (0.5, 4.5) \\ \\ \text{Gradient between two points} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {3 - 6 \over 2 - (-1)} \\ & = -1 \\ \\ \text{Gradient of perpendicular bisector} & = -1 \div -1 \phantom{000000} [m_1 \times m_2 = -1 \implies m_1 = -1 \div m_2] \\ & = 1 \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } & (0.5, 4.5), \\ 4.5 & = 0.5 + c \\ 4 & = c \\ \\ \text{Eqn of perp. bisector: } & y = x + 4 \phantom{0} \text{--- (1)} \\ \\ 2x & + 5y = -1 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x + 5(x + 4) & = -1 \\ 2x + 5x + 20 & = -1 \\ 7x & = -21 \\ x & = -3 \\ \\ \text{Substitute } & x = -3 \text{ into (1),} \\ y & = -3 + 4 \\ y & = 1 \\ \\ \therefore \text{Centre: } & (-3, 1) \end{align*}

The perpendicular bisector of chords $OA$ and $AB$ will meet at the centre

\begin{align*} \text{Midpoint of } OA & = \left({0 + 1 \over 2}, {0 + (-1) \over 2} \right) \\ & = \left( {1 \over 2}, -{1 \over 2}\right) \\ \\ \text{Gradient of } OA & = {-1 - 0 \over 1 - 0} \\ & = -1 \\ \\ \text{Gradient of perp. bisector} \times -1 & = -1 \\ \text{Gradient of perp. bisector} & = 1 \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } & \left({1 \over 2}, -{1 \over 2}\right), \\ -{1 \over 2} & = {1 \over 2} + c \\ -1 & = c \\ \\ \text{Eqn of perp. bisector of } OA : & \phantom{.} y = x - 1 \phantom{000} \text{--- (1)} \\ \\ \\ \text{Midpoint of } AB & = \left({1 + 4 \over 2}, {-1 + 0 \over 2} \right) \\ & = \left( {5 \over 2}, -{1 \over 2}\right) \\ \\ \text{Gradient of } AB & = {0 - (-1) \over 4 - 1} \\ & = {1 \over 3} \\ \\ \text{Gradient of perp. bisector} \times {1 \over 3} & = -1 \\ \text{Gradient of perp. bisector} & = {-1 \over {1 \over 3}} \\ & = -3 \\ \\ y & = mx + c \\ y & = -3x + c \\ \\ \text{Using } & \left({5 \over 2}, -{1 \over 2}\right), \\ -{1 \over 2} & = -3\left(5 \over 2\right) + c \\ -{1 \over 2} & = -{15 \over 2} + c \\ 7 & = c \\ \\ \text{Eqn of perp. bisector of } AB : & \phantom{.} y = -3x + 7 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x - 1 & = -3x + 7 \\ x + 3x & = 7 + 1 \\ 4x & = 8 \\ x & = {8 \over 4} \\ & = 2 \\ \\ \text{Substitute } & \text{into (1),} \\ y & = 2 - 1 \\ & = 1 \\ \\ \therefore \text{Centre: } & (2, 1) \end{align*}

\begin{align*} (x - 1)^2 + (y - 2)^2 & = 5 \\ \\ \text{Centre: } (1, 2), & \text{ Radius} = \sqrt{5} \text{ units} \\ \\ \text{Gradient of normal} & = \text{Gradient of } OP \\ & = { y_2 - y_1 \over x_2 - x_1} \\ & = { 4 - 2 \over 2 - 1} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & P(2, 4), \\ 4 & = 2(2) + c \\ 4 & = 4 + c \\ 0 & = c \\ \\ \text{Eqn of normal: } & y = 2x \end{align*}

\begin{align*} y & = -x + 6 \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of tangent} & = -1 \\ \\ \text{Gradient of normal} & = -1 \div -1 \phantom{000000} [m_1 \times m_2 = -1 \implies m_1 = -1 \div m_2] \\ & = 1 \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } & \text{centre } (0, 1), \\ 1 & = 0 + c \\ 1 & = c \\ \\ \text{Eqn of normal: } & y = x + 1 \phantom{0} \text{--- (1)} \\ \\ \text{Eqn of tangent: } & y = -x + 6 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x + 1 & = -x + 6 \\ 2x & = 5 \\ x & = 2.5 \\ \\ \text{Substitute } & x = 2.5 \text{ into (1),} \\ y & = 2.5 + 1 \\ y & = 3.5 \\ \\ \therefore & \phantom{.} A(2.5, 3.5) \end{align*}

\begin{align} [x - (-3)]^2 + (y - 2)^2 & = 2^2 \\ (x + 3)^2 + (y - 2)^2 & = 4 \end{align}

\begin{align} \text{Radius} & = 2 \text{ units} \\ \\ [x - (-2)]^2 + (y - 2)^2 & = 2^2 \\ (x + 2)^2 + (y - 2)^2 & = 4 \end{align}

\begin{align} \text{Radius} & = 6 \text{ units} \\ \\ \text{Centre: } & (6, -6) \\ \\ (x - 6)^2 + [y - (-6)]^2 & = 6^2 \\ (x - 6)^2 + (y + 6)^2 & = 36 \end{align}