(a)

\begin{align} (x - 0)^2 + (y - 1)^2 & = (4)^2 \\ (x)^2 + (y)^2 - 2(y)(1) + (1)^2 & = 16 \\ x^2 + y^2 - 2y + 1 - 16 & = 0 \\ x^2 + y^2 - 2y - 15 & = 0 \end{align}

(b)

Since the circle touches the $x$-axis and it's center is $(3, -2)$, the radius is $2$ units \begin{align} \text{Radius} & = 2 \text{ units} \\ \\ (x - 3)^2 + [y - (-2)]^2 & = 2^2 \\ (x)^2 - 2(x)(3) + (3)^2 + (y + 2)^2 & = 4 \\ x^2 - 6x + 9 + (y)^2 + 2(y)(2) + (2)^4 & = 4 \\ x^2 - 6x + 9 + y^2 + 4y + 4 & = 4 \\ x^2 + y^2 - 6x + 4y + 9 & = 0 \end{align}

(c)

Since the circle touches the $y$-axis and it's center is $(-3, 4)$, the radius is $3$ units \begin{align} \text{Radius} & = 3 \text{ units} \\ \\ [x - (-3)]^2 + (y - 4)^2 & = 3^2 \\ (x + 3)^2 + (y - 4)^2 & = 9 \\ (x)^2 + 2(x)(3) + (3)^2 + (y)^2 - 2(y)(4) + (4)^2 & = 9 \\ x^2 + 6x + 9 + y^2 - 8y + 16 & = 9 \\ x^2 + y^2 + 6x - 8y + 16 & = 0 \end{align}

(d)

Since the circle touches both axes and it's center is $(-2, 2)$, the radius is $2$ units \begin{align} \text{Radius} & = 2 \text{ units} \\ \\ [x - (-2)]^2 + (y - 2)^2 & = 2^2 \\ (x + 2)^2 + (y - 2)^2 & = 4 \\ (x)^2 + 2(x)(2) + (2)^2 + (y)^2 - 2(y)(2) + (2)^2 & = 4 \\ x^2 + 4x + 4 + y^2 - 4y + 4 & = 4 \\ x^2 + y^2 + 4x - 4y + 4 & = 0 \end{align}

(a)

\begin{align} x^2 + y^2 + 2x - 10y + 1 & = 0 \\ x^2 + y^2 + 2(1)(x) + 2(-5)(y) + 1 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = 1\\ f & = -5 \\ c & = 1 \\ \\ \therefore \text{Centre } & (-1, 5) \\ \\ \therefore \text{Radius} & = \sqrt{(1)^2 + (-5)^2 - 1} \\ & = 5 \text{ units} \end{align}

(b)

\begin{align} x^2 + y^2 - 4x - 6y + 4 & = 0 \\ x^2 + y^2 + 2(-2)(x) + 2(-3)(y) + 4 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = -2 \\ f & = -3 \\ c & = 4 \\ \\ \therefore \text{Centre } & (2, 3) \\ \\ \therefore \text{Radius} & = \sqrt{(-2)^2 + (-3)^2 - 4} \\ & = 3 \text{ units} \end{align}

\begin{align} (x - 2)^2 + [y - (-1)]^2 & = 5^2 \\ (x - 2)^2 + (y + 1)^2 & = 25 \\ (x)^2 - 2(x)(2) + (2)^2 + (y)^2 + 2(y)(1) + (1)^2 & = 25 \\ x^2 - 4x + 4 + y^2 + 2y + 1 & = 25 \\ x^2 + y^2 - 4x + 2y - 20 & = 0 \\ x^2 + y^2 + 2(-2)(x) + 2(1)(y) + (-20) & = 0 \\ \\ \text{Comparing with } x^2 + y^2 + 2gx & + 2fy + c = 0, \\ \therefore g & = - 2 \\ f & = 1 \\ c & = -20 \end{align}

(i)

\begin{align} x - 2y & = 6 \\ x & = 6 + 2y \phantom{000} \text{ --- (1)} \\ \\ y - 2x & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ y - 2(6 + 2y) & = 0 \\ y - 12 - 4y & = 0 \\ -3y & = 12 \\ y & = {12 \over -3} \\ & = - 4 \\ \\ \text{Substitute } & y = -4 \text{ into (1),} \\ x & = 6 + 2(-4) \\ & = -2 \\ \\ \therefore \text{Point of intersection } & (-2, - 4) \\ \\ \text{Radius} & = \sqrt{ (-2 - 0)^2 + (-4 - 0)^2 } \\ & = \sqrt{ 20 } \text{ units} \\ \\ (x - 0)^2 + (y - 0)^2 & = (\sqrt{20})^2 \\ (x)^2 + (y)^2 & = 20 \\ x^2 + y^2 & = 20 \\ x^2 + y^2 - 20 & = 0 \end{align}

(ii)

\begin{align} \text{When } & x = 2, \\ (2)^2 + y^2 - 20 & = 0 \\ 4 + y^2 - 20 & = 0 \\ y^2 - 16 & = 0 \\ y^2 & = 16 \\ y & = \pm \sqrt{16} \\ & = \pm 4 \phantom{000} (\ne 6) \\ \\ \therefore (2, 6) \text{ does not} & \text{ lie on the circle} \end{align}

Alternative method for (ii): Find the distance between the centre $(0, 0)$ and the point $(2, 6)$. If the distance is smaller (or larger) than the radius $\sqrt{20}$, the point does not lie on the circle.

(i)

\begin{align} 12x - 5y & = -11 \\ 12x & = 5y - 11 \\ x & = {5y - 11 \over 12} \phantom{000} \text{ --- (1)} \\ \\ x^2 + y^2 - 6x + 2y - 15 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left(5y - 11 \over 12\right)^2 + y^2 - 6\left(5y - 11 \over 12\right) + 2y - 15 & = 0 \\ {(5y - 11)^2 \over 144} + y^2 - {6(5y - 11) \over 12} + 2y - 15 & = 0 \\ (5y - 11)^2 + 144y^2 - 72(5y - 11) + 288y - 2160 & = 0 \phantom{00000} \text{[Multiply both sides by 144]} \\ (5y)^2 - 2(5y)(11) + (11)^2 + 144y^2 - 360y + 792 + 288y - 2160 & = 0 \\ 25y^2 - 110y + 121 + 144y^2 - 72y - 1368 & = 0 \\ 169y^2 - 182y - 1247 & = 0 \\ (13y - 43)(13y + 29) & = 0 \\ \\ 13y - 43 = 0 \phantom{000}&\text{or} \phantom{00} 13y + 29 = 0 \\ 13y = 43 \phantom{00} & \phantom{or00+ 29} 13y = -29 \\ y = {43 \over 13} \phantom{0} & \phantom{or00+2913} y = -{29 \over 13} \\ \\ \\ \text{Substitute } y & = {43 \over 13} \text{ into (1),} \\ x & = {5\left(43 \over 13\right) - 11 \over 12} \\ & = {6 \over 13} \\ \\ \therefore & \phantom{0} \left({6 \over 13}, {43 \over 13}\right) \\ \\ \text{Substitute } y & = -{29 \over 13} \text{ into (2),} \\ x & = {5 \left(-{29 \over 13}\right) - 11 \over 12} \\ & = -{24 \over 13} \\ \\ \therefore & \phantom{0} \left(-{24 \over 13}, -{29 \over 13}\right) \end{align}

(ii)

\begin{align} \text{Length of } AB & = \sqrt{ \left[ {6 \over 13} - \left(-{24 \over 13}\right) \right]^2 + \left[ {43 \over 13} - \left( - {29 \over 13} \right) \right]^2 } \\ & = 6 \text{ units} \end{align}

(i)

\begin{align} y & = 2x \phantom{000} \text{ --- (1)} \\ \\ x^2 + y^2 + 4x - 6y + 3 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (2x)^2 + 4x - 6(2x) + 3 & = 0 \\ x^2 + 4x^2 + 4x - 12x + 3 & = 0 \\ 5x^2 - 8x + 3 & = 0 \\ (5x - 3)(x - 1) & = 0 \\ \\ 5x - 3 = 0 \phantom{00}&\text{or}\phantom{00} x - 1 = 0 \\ 5x = 3 \phantom{00}&\phantom{or00-1} x = 1 \\ x = {3 \over 5} \phantom{-} & \\ \\ \text{Substitute } x & = {3 \over 5} \text{ into (1),} \\ y & = 2\left(3 \over 5\right) \\ & = {6 \over 5} \\ \\ \therefore & \phantom{0} \left({3 \over 5}, {6 \over 5} \right) \\ \\ \text{Substitute } x & = 1 \text{ into (1),} \\ y & = 2(1) \\ & = 2 \\ \\ \therefore & \phantom{0} (1, 2) \end{align}

(ii)

\begin{align} \text{Mid-point of } PQ & = \left( {1 + {3 \over 5} \over 2}, {2 + {6 \over 5} \over 2} \right) \\ & = \left({4 \over 5}, {8 \over 5} \right) \\ \\ \text{Gradient of } PQ & = {2 - {6 \over 5} \over 1 - {3 \over 5} } \\ & = 2 \\ \\ \text{Gradient of perp. bisector} \times 2 & = -1 \\ \text{Gradient of perp. bisector} & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & \left({4 \over 5}, {8 \over 5} \right), \\ {8 \over 5} & = -{1 \over 2}\left(4 \over 5\right) + c \\ {8 \over 5} & = -{2 \over 5} + c \\ 2 & = c \\ \\ \text{Eqn of perp. bisector of } PQ: & \phantom{.} y = -{1 \over 2}x + 2 \end{align}

(i)

\begin{align} x^2 + y^2 - 2x + 8y - 23 & = 0 \\ x^2 + y^2 + 2(-1)(x) + 2(4)(y) - 23 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = - 1 \\ f & = 4 \\ c & = -23 \\ \\ \therefore \text{Centre } & (1, -4) \\ \\ \text{Radius} & = \sqrt{g^2 + f^2 - c} \\ & = \sqrt{(-1)^2 + (4)^2 - (-23)} \\ & = \sqrt{40} \\ & = \sqrt{4}\sqrt{10} \\ & = 2\sqrt{10} \text{ units} \end{align}

(ii)

\begin{align} \text{Let the} & \text{ centre (1, -4) be } C. \\ \text{Let the} & \text{ point (3, -10) be } D. \end{align}

$CD$ is the radius of the circle and it is perpendicular to the tangent $AB$ (Circle property from E Maths) \begin{align} \text{Gradient of } CD & = {-4 -(-10) \over 1 - 3} \\ & = - 3 \\ \\ \text{Gradient of } AB \times -3 & = -1 \\ \text{Gradient of } AB & = {-1 \over -3} \\ & = {1 \over 3} \\ \\ y & = mx + c \\ y & = {1 \over 3}x + c \\ \\ \text{Using } & D(3, -10), \\ -10 & = {1 \over 3}(3) + c \\ -10 & = 1 + c \\ -11 & = c \\ \\ & \phantom{...} y = {1 \over 3}x - 11 \\ \text{Eqn of line } AB: & \phantom{...} 3y = x - 33 \end{align}

(i)

\begin{align} \text{Mid-point of } AB & = \left( {-2 + 4 \over 2}, {0 + 0 \over 2} \right) \\ & = (1, 0) \\ \\ \text{Gradient of } AB & = {0 - 0 \over -2 - 4} \\ & = 0 \\ \\ \text{Since } AB \text{ is a horizontal line, the perp. } & \text{bisector is a vertical line passing through (1, 0).} \\ \\ \therefore \text{Equation of perp. bisector:} & \phantom{00} x = 1 \end{align}

(ii)

(iii)

\begin{align} \text{Radius} & = \text{Length of } AC \\ & = \sqrt{ (-2 - 1)^2 + (0 - 4)^2 } \\ & = \sqrt{ 25 } \\ & = 5 \text{ units} \\ \\ (x - 1)^2 + (y - 4)^2 & = 5^2 \\ (x)^2 - 2(x)(1) + (1)^2 + (y)^2 - 2(y)(4) + (4)^2 & = 25 \\ x^2 - 2x + 1 + y^2 - 8y + 16 & = 25 \\ x^2 + y^2 - 2x - 8y + 1 + 16 - 25 & = 0 \\ x^2 + y^2 - 2x - 8y - 8 & = 0 \end{align}

(i)

\begin{align} \text{When } & y = 0, \\ x^2 + (0)^2 - 4x + 6(0) - 12 & = 0 \\ x^2 - 4x - 12 & = 0 \\ (x - 6)(x + 2) & = 0 \\ \\ x - 6 = 0 \phantom{00}&\text{or}\phantom{00} x + 2 = 0 \\ x = 6 \phantom{00}&\phantom{or00+2} x = - 2 \\ \\ \therefore A(-2, 0), & \phantom{0} B(6,0) \\ \\ \text{Length of } AB & = \sqrt{ (-2 - 6)^2 + (0 - 0)^2 } \\ & = 8 \text{ units} \end{align}

(ii) Since AD is the diameter, the midpoint of AD is the centre of the circle

\begin{align} x^2 + y^2 - 4x + 6y - 12 & = 0 \\ x^2 + y^2 + 2(-2)(x) + 2(3)(y) - 12 & = 0 \\ \\ \text{Comparing with } & x^2 + y^2 + 2gx + 2fy + c = 0, \\ g & = - 2 \\ f & = 3 \\ c & = -12 \\ \\ \therefore \text{Centre } & (2, - 3) \\ \\ \text{Let coordinates of } & D \text{ be } (a, b). \\ \\ \text{Mid-point of } AD & = \left( {a + (-2) \over 2}, {b + 0 \over 2} \right) \\ (2, - 3) & = \left( {a - 2 \over 2}, {b \over 2} \right) \\ \\ \\ \text{Comparing the } & x \text{-coordinate,} \\ 2 & = {a - 2 \over 2} \\ 2(2) & = a - 2 \\ 4 & = a - 2 \\ 4 + 2 & = a \\ 6 & = a \\ \\ \text{Comparing the } & y \text{-coordinate,} \\ -3 & = {b \over 2} \\ 2(-3) & = b \\ -6 & = b \\ \\ \therefore & \phantom{0} D (6, -6) \end{align}

(i)

Since both circles touches the axes and lie on the line $y = x$, one circle is in the first quadrant and the other in the third quadrant.

\begin{align} \text{Centre of } C_1 \text{ is }& (5, 5) \\ \\ (x - 5)^2 + (y - 5)^2 & = 5^2 \\ (x)^2 - 2(x)(5) + (5)^2 + (y)^2 - 2(y)(5) + (5)^2 & = 25 \\ x^2 - 10x + 25 + y^2 - 10y + 25 & = 25 \\ x^2 + y^2 - 10x - 10y + 25 + 25 - 25 & = 0 \\ \\ \text{Equation of } C_1: x^2 + y^2 - 10x - 10y + 25 & = 0 \\ \\ \\ \text{Centre of } C_2 \text{ is } & (-5, -5) \\ \\ [x - (-5)]^2 + [y - (-5)]^2 & = 5^2 \\ (x + 5)^2 + (y + 5)^2 & = 25 \\ (x)^2 + 2(x)(5) + (5)^2 + (y)^2 + 2(y)(5) + (5)^2 & = 25 \\ x^2 + 10x + 25 + y^2 + 10y + 25 & = 25 \\ x^2 + y^2 + 10x + 10y + 25 + 25 - 25 & = 0 \\ \\ \text{Equation of } C_2: x^2 + y^2 + 10x + 10y + 25 & = 0 \end{align}

(ii)

Since circle $C_3$ passes through the centres of $C_1$ and $C_2$, the centre of $C_3$ is equidistant from both centres. Thus, the perpendicular bisector ($y = -x$) of the centres of $C_1$ and $C_2$ passes through the point $(0,0)$.

Constructing a circle with centre $(0,0)$ and radius $6$ units, the intersection points of the circle with the line $y = -x$ are the possible coordinates of the centre of $C_3$.

\begin{align} y & = -x \phantom{000} \text{ --- (1)} \\ \\ (x - 0)^2 + (y - 0)^2 & = 6^2 \\ (x)^2 + (y)^2 & = 36 \\ x^2 + y^2 & = 36 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (-x)^2 & = 36 \\ x^2 + x^2 & = 36 \\ 2x^2 & = 36 \\ x^2 & = 18 \\ x & = \pm \sqrt{18} \\ & = \pm \sqrt{9} \sqrt{2} \\ & = \pm 3\sqrt{2} \\ \\ \text{Substitute } x & = 3\sqrt{2} \text{ into (1),} \\ y & = -(3\sqrt{2}) \\ & = -3\sqrt{2} \\ \\ \therefore & \phantom{0} (3\sqrt{2}, -3\sqrt{2}) \\ \\ \text{Substitute } x & = -3\sqrt{2} \text{ into (1),} \\ \\ y & = -(-3\sqrt{2}) \\ & = 3\sqrt{2} \\ \\ \therefore & \phantom{0} (-3\sqrt{2}, 3\sqrt{2}) \end{align}

\begin{align} \text{Radius} & = \text{Distance between centres of } C_1 \text{ and } C_3 \\ & = \sqrt{ (5 - 3\sqrt{2})^2 + [5 - (-3\sqrt{2})]^2 } \\ & = \sqrt{ (5)^2 - 2(5)(3\sqrt{2}) + (3\sqrt{2})^2) + (5)^2 + 2(5)(3\sqrt{2}) + (3\sqrt{2})^2 } \\ & = \sqrt{ 25 - 30\sqrt{2} + 18 + 25 + 30\sqrt{2} + 18 } \\ & = \sqrt{ 86 } \end{align} \begin{align} \text{Using } (3\sqrt{2}, - 3\sqrt{2}) &, \\ (x - 3\sqrt{2})^2 + [y - (-3\sqrt{2})]^2 & = (\sqrt{86})^2 \\ (x - 3\sqrt{2})^2 + (y + 3\sqrt{2})^2 & = 86 \\ x^2 - 6\sqrt{2}x + 18 + y^2 + 6\sqrt{2}y + 18 & = 86 \\ \\ \therefore x^2 + y^2 - 6\sqrt{2}x + 6\sqrt{2}y - 50 & = 0 \\ \\ \\ \text{Using } (-3\sqrt{2}, 3\sqrt{2}) &, \\ [x - (-3\sqrt{2})]^2 + (y - 3\sqrt{2})^2 & = (\sqrt{86})^2 \\ (x + 3\sqrt{2})^2 + (y - 3\sqrt{2})^2 & = 86 \\ x^2 + 6\sqrt{2}x + 18 + y^2 - 6\sqrt{2}y + 18 & = 86 \\ \\ \therefore x^2 + y^2 + 6\sqrt{2}x - 6\sqrt{2}y - 50 & = 0 \end{align}

(i) The tangent at P is perpendicular to the line passing through P and the centre of the circle (Circle property from E Maths)

\begin{align} x + 2y & = -5 \\ 2y & = -x - 5 \\ y & = -{1 \over 2}x - {5 \over 2} \\ \\ \text{Comparing with } & y = mx + c, \\ \text{Gradient, } m & = -{1 \over 2} \\ \\ \text{Let centre of the } & \text{circle be } C. \\ \\ \text{Gradient of } PC \times -{1 \over 2} & = -1 \\ \text{Gradient of } PC & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & C(-1, 3), \\ 3 & = 2(-1) + c \\ 3 & = -2 + c \\ 5 & = c \\ \\ \text{Equation of }PC: \phantom{0} y & = 2x + 5 \phantom{000} \text{ --- (1)} \\ \\ \text{Equation of tangent at } P: \phantom{0} x + 2y & = -5 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x + 2(2x + 5) & = -5 \\ x + 4x + 10 & = -5 \\ 5x & = -15 \\ x & = {-15 \over 5} \\ & = -3 \\ \\ \text{Substitute } x & = -3 \text{ into } (1), \\ y & = 2(-3) + 5 \\ & = -1 \\ \\ \therefore & \phantom{0} P (-3, -1) \end{align}

(ii)

\begin{align} \text{Radius} & = \text{Length of } PC \\ & = \sqrt{ [-3 - (-1)]^2 + (-1 - 3)^2 } \\ & = \sqrt{20} \\ \\ [x - (-1)]^2 + (y - 3)^2 & = (\sqrt{20})^2 \\ (x + 1)^2 + (y - 3)^2 & = 20 \\ (x)^2 + 2(x)(1) + (1)^2 + (y)^2 - 2(y)(3) + (3)^2 & = 20 \\ x^2 + 2x + 1 + y^2 - 6y + 9 & = 20 \\ x^2 + y^2 + 2x - 6y + 1 + 9 - 20 & = 0 \\ \\ \therefore x^2 + y^2 + 2x - 6y - 10 & = 0 \end{align}

\begin{align} y & = 3x - 7 \phantom{000} \text{ --- (1)} \\ \\ x^2 + y^2 - 4x - 8y - 5 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (3x - 7)^2 - 4x - 8(3x - 7) - 5 & = 0 \\ x^2 + (3x)^2 - 2(3x)(7) + (7)^2 - 4x - 24x + 56 - 5 & = 0 \\ x^2 + 9x^2 - 42x + 49 - 28x + 51 & = 0 \\ 10x^2 - 70x + 100 & = 0 \\ x^2 - 7x + 10 & = 0 \\ (x - 2)(x - 5) & = 0 \\ \\ x - 2 = 0 \phantom{00}&\text{or} \phantom{00} x - 5 = 0 \\ x = 2 \phantom{00}&\phantom{or00-5} x = 5 \\ \\ \text{Substitute } x & = 2 \text{ into (1),} \\ y & = 3(2) - 7 \\ & = -1 \\ \\ \therefore & \phantom{0} A(2, -1) \\ \\ \text{Substitute } x & = 5 \text{ into (1),} \\ y & = 3(5) - 7 \\ & = 8 \\ \\ \therefore & \phantom{0} B(5, 8) \\ \\ \\ x^2 + y^2 - 4x - 8y - 5 & = 0 \\ x^2 + y^2 + 2(-2)(x) + 2(-4)(y) - 5 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fx + c = 0, \\ g & = -2 \\ f & = -4 \\ \\ \therefore & \phantom{0} C(2, 4) \\ \end{align} \begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 5 & 2 & 2 & 5 \\ 8 & 4 & -1 & 8 \end{matrix} \right| \\ & = {1 \over 2} [5 \times 4 + 2 \times (-1) + 2 \times 8 - 8 \times 2 - 4 \times 2 - (-1) \times 5] \\ & = {1 \over 2} (15) \\ & = 7.5 \text{ sq. units} \end{align}

\begin{align} x^2 + y^2 + 4x + 6y - 12 & = 0 \\ x^2 + y^2 + 2(2)(x) + 2(3)(y) - 12 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = 2 \\ f & = 3 \\ c & = -12 \\ \\ \therefore \text{Centre } & (-2, -3) \\ \\ \text{Radius} & = \sqrt{g^2 + f^2 - c} \\ & = \sqrt{ (2)^2 + (3)^2 - (-12) } \\ & = \sqrt{ 25 } \\ & = 5 \text{ units} \\ \\ \text{Distance between centre and } (-3, -1) & = \sqrt{ [-3 - (-2)]^2 + [-1 - (-3)]^2 } \\ & = \sqrt{ 5 } \text{ units } < 5 \text{ units (Radius)} \\ \\ \\ \therefore \text{Point (-3, -1) lies inside the circle, } & \text{thus any line passing through (-3, -1) cannot be tangent to circle} \end{align}

Extra: Since (-3, -1) lies inside the circle, any line that passes through the point (-3, -1) will meet the circle at two points

The line AC (i.e. the radius of the circle) is perpendicular to the tangent at the point A (Circle property from E Maths)

\begin{align} x^2 + y^2 + 2x - 2y - 3 & = 0 \\ x^2 + y^2 + 2(1)(x) + 2(-1)(y) - 3 & = 0 \\ \\ \text{Compare with } x^2 + y^2 + 2gx & + 2fy + c = 0, \\ g & = 1 \\ f & = -1 \\ \\ \therefore \text{Centre, }C & (-1, 1) \\ \\ \text{Let point (1,2) } & \text{be } A. \\ \\ \text{Gradient of } AC & = {1 - 2 \over -1 - 1} \\ & = {1 \over 2} \\ \\ \text{Gradient of tangent} \times {1 \over 2} & = -1 \\ \text{Gradient of tangent} & = -1 \div {1 \over 2} \\ & = -2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & A(1, 2), \\ 2 & = -2(1) + c \\ 2 & = -2 + c \\ 4 & = c \\ \\ \text{Eqn of tangent at } A(1,2): & \phantom{.} y = -2x + 4 \end{align}

(i)

\begin{align} \text{Let } & \text{coordinates of } P \text{ be } (x, y). \\ \\ PA & = \sqrt{ [x - (-a)]^2 + (y - 0)^2 } \\ & = \sqrt{ (x + a)^2 + y^2 } \\ & = \sqrt{ x^2 + 2(x)(a) + a^2 + y^2} \\ & = \sqrt{ x^2 + 2ax + a^2 + y^2 } \\ \\ PB & = \sqrt{ (x - a)^2 + (y - 0)^2 }\\ & = \sqrt{ x^2 - 2(x)(a) + a^2 + y^2 } \\ & = \sqrt{ x^2 - 2ax + a^2 + y^2 } \\ \\ \text{Since } & PA = 2PB, \\ \sqrt{x^2 + 2ax + a^2 + y^2} & = 2\sqrt{x^2 - 2ax + a^2 + y^2} \\ \left(\sqrt{x^2 + 2ax + a^2 + y^2} \right)^2 & = \left(2\sqrt{x^2 - 2ax + a^2 + y^2}\right)^2 \\ x^2 + 2ax + a^2 + y^2 & = 4(x^2 - 2ax + a^2 + y^2) \\ x^2 + 2ax + a^2 + y^2 & = 4x^2 - 8ax + 4a^2 + 4y^2 \\ 0 & = 3x^2 - 10ax + 3a^2 + 3y^2 \\ 0 & = x^2 - {10 \over 3}ax + a^2 + y^2 \\ 0 & = x^2 + y^2 - {10 \over 3}ax + a^2 \\ \\ x^2 + y^2 - {10 \over 3}ax + a^2 & = 0 \\ x^2 + y^2 + 2\left(-{5 \over 3}a\right)(x) + 2(0)(y) + a^2 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = -{5 \over 3}a \\ f & = 0 \\ c & = a^2 \\ \\ \therefore \text{Equation of path } & \text{is a circle} \end{align}

(ii)

\begin{align} \text{Centre } & \left( {5 \over 3}a, 0 \right) \\ \\ \text{Radius} & = \sqrt{ g^2 + f^2 - c } \\ & = \sqrt{ \left({5 \over 3}a\right)^2 + (0)^2 - a^2 } \\ & = \sqrt{ {25 \over 9}a^2 - a^2 } \\ & = \sqrt{ {16 \over 9}a^2 } \\ & = {4 \over 3}a \text{ units} \end{align}

(i)

\begin{align} \text{Radius of } C_1 & = \text{Length of } PQ \\ & = \sqrt{ (2 - 6)^2 + (-2 - 0)^2 } \\ & = \sqrt{20} \text{ units} \\ \\ (x - 2)^2 + [y - (-2)]^2 & = (\sqrt{20})^2 \\ (x - 2)^2 + (y + 2)^2 & = 20 \\ x^2 - 2(x)(2) + 2^2 + y^2 + 2(y)(2) + 2^2 & = 20 \\ x^2 - 4x + 4 + y^2 + 4y + 4 & = 20 \\ x^2 + y^2 - 4x + 4y - 12 & = 0 \end{align}

(ii)

\begin{align} \text{Mid-point of } PQ & = \left( {2 + 6 \over 2}, {-2 + 0 \over 2} \right) \\ & = (4, -1) \\ \\ \text{Gradient of } PQ & = {-2 - 0 \over 2 - 6} \\ & = {1 \over 2} \\ \\ \text{Gradient of perp. bisector} \times {1 \over 2} & = -1 \\ \text{Gradient of perp. bisector} & = -1 \div {1 \over 2} \\ & = -2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & (4, -1), \\ -1 & = -2(4) + c \\ -1 & = -8 + c \\ 7 & = c \\ \\ \text{Eqn of perp. bisector of } & PQ: \phantom{0} y = -2x + 7 \end{align}

(iii)

(iv)

\begin{align} (x - 6)^2 + [y - (-5)]^2 & = 5^2 \\ (x - 6)^2 + (y + 5)^2 & = 25 \\ x^2 - 2(x)(6) + 6^2 + y^2 + 2(y)(5) + 5^2 & = 25 \\ x^2 - 12x + 36 + y^2 + 10y + 25 & = 25 \\ x^2 + y^2 - 12x + 10y + 36 + 25 - 25 & = 0 \\ \\ x^2 + y^2 - 12x + 10y + 36 & = 0 \end{align}

(i)

\begin{align} x^2 + y^2 - 4x - 4y & = 1 \phantom{000} \text{ --- (1)} \\ \\ y & = x + c \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ x^2 + (x + c)^2 - 4x - 4(x + c) & = 1 \\ x^2 + x^2 + 2(x)(c) + c^2 - 4x - 4x - 4c & = 1 \\ 2x^2 + 2cx - 8x + c^2 - 4c - 1 & = 0 \\ x^2 + cx - 4x + {c^2 - 4c - 1 \over 2} & = 0 \\ \\ x^2 + (c - 4)x + {c^2 - 4c - 1 \over 2} & = 0 \phantom{0} \text{ (Shown)} \end{align}

(ii)

(iii)

\begin{align} \text{Let } x \text{ be the }& x\text{-coordinate of the point traced out by } M. \\ \\ x & = {4 - c \over 2} \\ 2x & = 4 - c \\ c & = 4 - 2x \phantom{000} \text{ --- (3)} \\ \\ \text{Let } y \text{ be the }& y\text{-coordinate of the point traced out by } M. \\ \\ y & = {4 + c \over 2} \\ 2y & = 4 + c \phantom{000} \text{ --- (4)} \\ \\ \text{Substitute } & \text{(3) into (4),} \\ 2y & = 4 + (4 - 2x) \\ 2y & = 8 - 2x \\ y & = 4 - x \end{align}

We need the circle property from E maths: Tangent perpendicular to radius

\begin{align} (x - 2)^2 + (y + 1)^2 & = 13 \\ (x - 2)^2 + [y - (-1)]^2 & = (\sqrt{13})^2 \\ \\ \text{Centre, } & C(2, -1) \\ \\ \text{Radius} & = \sqrt{13} \text{ units} \\ \\ CT & = \sqrt{13} \text{ units} \\ \\ CP & = \sqrt{ (2 - 9)^2 + (-1 - 2)^2 } \\ & = \sqrt{58} \text{ units} \\ \\ CP^2 & = CT^2 + PT^2 \\ (\sqrt{58})^2 & = (\sqrt{13})^2 + PT^2 \\ 58 & = 13 + PT^2 \\ 45 & = PT^2 \\ \sqrt{45} & = PT \\ \sqrt{9} \sqrt{5} & = PT \\ \\ PT & = 3 \sqrt{5} \text{ units} \end{align}

\begin{align} x & = a(y - k)^2 + h \\ \\ \text{Let } & x = 0, \\ 0 & = a(y - k)^2 + h \\ 0 & = a[(y)^2 - 2(y)(k) + (k)^2] + h \\ 0 & = a(y^2 - 2ky + k^2) + h \\ 0 & = ay^2 - 2aky + ak^2 + h \\ \\ b^2 - 4ac & = (-2ak)^2 - 4(a)(ak^2+ h) \\ & = 4a^2 k^2 - 4a(ak^2 + h) \\ & = 4a^2 k^2 - 4a^2k^2 - 4ah \\ & = -4ah \\ \\ \text{Since } a > 0 & \text{ and } h < 0, \phantom{.} b^2 - 4ac > 0 \\ \\ \therefore \text{Parabola } & \text{has two } y \text{-intercepts} \end{align}

The perpendicular bisectors of chords AC and AB will meet at the centre of the circle

\begin{align} \text{Midpoint of } AC & = \left({3 + (-1) \over 2}, {1 + 3 \over 2}\right) \\ & = (1, 2) \\ \\ \text{Gradient of } AC & = {3 - 1 \over -1 - 3} \\ & = -{1 \over 2} \\ \\ \text{Gradient of perp. bisector} \times -{1 \over 2} & = -1 \\ \text{Gradient of perp. bisector} & = -1 \div -{1 \over 2} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & (1, 2), \\ 2 & = 2(1) + c \\ 2 & = 2 + c \\ 0 & = c \\ \\ \text{Eqn of perp. bisector of } & AC: \phantom{.} y = 2x \phantom{000} \text{--- (1)} \\ \\ \\ \text{Midpoint of } AB & = \left({3 + 4 \over 2}, {1 + 5 \over 2}\right) \\ & = \left({7 \over 2}, 3 \right) \\ \\ \text{Gradient of } AB & = {5 - 1 \over 4 - 3} \\ & = 4 \\ \\ \text{Gradient of perp. bisector} \times 4 & = -1 \\ \text{Gradient of perp. bisector} & = {-1 \over 4} \\ & = -{1 \over 4} \\ \\ y & = mx + c \\ y & = -{1 \over 4}x + c \\ \\ \text{Using } & \left({7 \over 2}, 3 \right), \\ 3 & = -{1 \over 4}\left(7 \over 2\right) + c \\ 3 & = -{7 \over 8} + c \\ {31 \over 8} & = c \\ \\ \text{Eqn of perp. bisector of } & AB: \phantom{.} y = -{1 \over 4}x + {31 \over 8} \phantom{000} \text{--- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x & = -{1 \over 4}x + {31 \over 8} \\ {9 \over 4}x & = {31 \over 8} \\ x & = {31 \over 8} \div {9 \over 4} \\ & = {31 \over 18} \\ \\ \text{Substitute } & x = {31 \over 18} \text{ into (1),} \\ y & = 2 \left(31 \over 18\right) \\ & = {31 \over 9} \\ \\ \text{Centre: } & \left({31 \over 18}, {31 \over 9}\right) \\ \\ \text{Radius} & = \text{Distance between } A(3, 1) \text{ and centre } \left({31 \over 18}, {31 \over 9}\right) \\ & = \sqrt{ \left(3 - {31 \over 18} \right)^2 + \left(1 - {31 \over 9}\right)^2 } \\ & = \sqrt{ 2465 \over 324 } \\ \\ \\ \left(x - {31 \over 18}\right)^2 & + \left(y - {31 \over 9}\right)^2 = {2465 \over 324} \end{align}