\begin{align*} 2x^2 + 2y^2 & + 8x - 12y + 8 = 0 \\ x^2 + y^2 & + 4x - 6y + 4 = 0 \\ \\ 2g & = 4 \\ g & = {4 \over 2} = 2 \\ \\ 2f & = -6 \\ f & = {-6 \over 2} = - 3 \\ \\ c & = 4 \\ \\ \therefore \text{Coordinates} & \text{ of centre is } (-2, 3) \\ \\ \text{Radius} & = \sqrt{2^2 + (-3)^2 - 4} \\ & = 3 \text{ units} \end{align*}

\begin{align*} 2x^2 + 2y^2 + 8x - 12y + 8 & = 0 \\ x^2 + y^2 + 4x - 6y + 4 & = 0 \\ x^2 + 4x + y^2 - 6y + 4 & = 0 \\ \left(x + {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 + y^2 - 6y + 4 & = 0 \phantom{00000} [\text{Complete the square for } x] \\ (x + 2)^2 - 4 + y^2 - 6y + 4 & = 0 \\ (x + 2)^2 + y^2 - 6y & = 0 \\ (x + 2)^2 + \left(y - {6 \over 2}\right)^2 - \left(6 \over 2\right)^2 & = 0 \phantom{00000} [\text{Complete the square for } y ] \\ (x + 2)^2 + (y - 3)^2 - 9 & = 0 \\ (x + 2)^2 + (y - 3)^2 & = 9 \\ (x + 2)^2 + (y - 3)^2 & = 3^2 \\ \\ \therefore \text{Coordinates} & \text{ of centre is } (-2, 3) \\ \\ \text{Radius} & = 3 \text{ units} \end{align*}

\begin{align*} \text{Let centre of circle} & \text{ be } (a, b) \\ \\ (x - a)^2 + (y - b)^2 & = 4^2 \\ \\ \text{Since circle passes thr} & \text{ough } O(0,0), \\ (0 - a)^2 + (0 - b)^2 & = 4^2 \\ (-a)^2 + (-b)^2 & = 16 \\ a^2 + b^2 & = 16 \phantom{0} \text{--- (1)} \\ \\ \text{Since line } y = 3x - 4 \text{ pa} & \text{sses through } (a, b), \\ b & = 3a - 4 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute (1) } & \text{into (2),} \\ a^2 + (3a - 4)^2 & = 16 \\ a^2 + \underbrace{ (3a)^2 - 2(3a)(4) + (4)^2 }_{(a - b)^2 = a^2 - 2ab + b^2} & = 16 \\ a^2 + 9a^2 - 24a + 16 & = 16 \\ 10a^2 - 24a & = 0 \\ 2a (5a - 12) & = 0 \end{align*} \begin{align*} 2a & = 0 && \text{ or } & 5a - 12 & = 0 \\ a & = 0 \text{ (N.A.)} &&& 5a & = 12 \\ & &&& a & = {12 \over 5} \\ & &&& a & = 2.4 \\ \\ & &&& \text{Substitute } & \text{into (2),} \\ & &&& b & = 3(2.4) - 4 \\ & &&& b & = 3.2 \\ \\ & &&& \therefore & \phantom{.} (2.4, 3.2) \end{align*}

\begin{align*} x^2 + y^2 - 10 & = 0 \\ x^2 + y^2 & = 10 \\ (x - 0)^2 + (y - 0)^2 & = (\sqrt{10})^2 \\ \\ \text{Centre: } (0, 0), & \text{ Radius} = \sqrt{10} \text{ units} \\ \\ \text{Distance between }(0, 0) \text{ and } (2, 2) & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ (2 - 0)^2 + (2 - 0)^2 } \\ & = \sqrt{8} \text{ units} < \text{Radius} \\ \\ \therefore (2, 2) & \text{ lies inside the circle} \end{align*}

\begin{align*} \text{Length of } AB & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ (3 - 7)^2 + (4 - 6)^2 } \\ & = \sqrt{20} \\ & = \sqrt{4} \times \sqrt{5} \phantom{000000} [ \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} ] \\ & = 2 \sqrt{5} \text{ units} \\ \\ \text{Radius} & = { 2 \sqrt{5} \over 2 } \\ & = \sqrt{5} \text{ units} \\ \\ \text{Midpoint of } AB & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( { 7 + 3 \over 2}, {6 + 4 \over 2} \right) \\ & = (5, 5) \\ \\ \implies \text{Centre: } & (5, 5) \\ \\ \\ \text{Eqn: } (x - 5)^2 & + (y - 5)^2 = (\sqrt{5})^2 \\ (x - 5)^2 & + (y - 5)^2 = 5 \end{align*}

\begin{align*} (x + 1)^2 & + (y - 2)^2 = 16 \phantom{0} \text{--- (1)} \\ \\ \text{Centre: } & (-1, 2), \text{ Radius} = \sqrt{16} = 4 \text{ units} \\ \\ y & = -x + 5 \phantom{0} \text{--- (2)} \\ \\ \text{Substi} & \text{tute (2) into (1),} \end{align*} \begin{align*} (x + 1)^2 + (- x + 5 - 2)^2 & = 16 \\ (x + 1)^2 + (3 - x)^2 & = 16 \\ \underbrace{ (x)^2 + 2(x)(1) + (1)^2}_{(a + b)^2 = a^2 + 2ab + b^2} + \underbrace{ (3)^2 - 2(3)(x) + (x)^2}_{(a - b)^2 = a^2 - 2ab + b^2 } & = 16 \\ x^2 + 2x + 1 + 9 - 6x + x^2 & = 16 \\ 2x^2 - 4x + 10 & = 16 \\ 2x^2 - 4x - 6 & = 0 \\ x^2 - 2x - 3 & = 0 \\ (x - 3)(x + 1) & = 0 \end{align*} \begin{align*} x - 3 & = 0 && \text{ or } & x + 1 & = 0 \\ x & = 3 &&& x & = -1 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = -3 + 5 &&& y & = -(-1) + 5 \\ y & = 2 &&& y & = 6 \\ \\ \therefore & \phantom{.} (3, 2) &&& \therefore & \phantom{.} (-1, 6) \end{align*}
\begin{align*} \text{Length of } AB & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ (-1 - 3)^2 + (6 - 2)^2 } \\ & = \sqrt{ 32 } \text{ units} \\ & = 5.657 \text{ units} \\ \\ \text{Diameter of circle} & = 4 \times 2 \\ & = 8 \text{ units} \\ \\ \text{Since } AB < \text{Diameter}, & \phantom{.} AB \text{ does not pass through centre of circle} \end{align*}