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# Equation Of A Tangent To A Circle

## 7.3 Equation of a tangent to a circle (EMCHW)

1. On a suitable system of axes, draw the circle $$x^{2} + y^{2} = 20$$ with centre at $$O(0;0)$$.
2. Plot the point $$T(2;4)$$.
3. Plot the point $$P(0;5)$$. Draw $$PT$$ and extend the line so that is cuts the positive $$x$$-axis.
4. Measure $$O\hat{T}P$$.
5. Determine the gradient of the radius $$OT$$.
6. Determine the gradient of $$PT$$.
7. Prove that $$PT \perp OT$$.
8. Plot the point $$S(2;-4)$$ and join $$OS$$.
9. Draw a tangent to the circle at $$S$$.
10. Measure the angle between $$OS$$ and the tangent line at $$S$$.
11. Make a conjecture about the angle between the radius and the tangent to a circle at a point on the circle.
12. Complete the sentence: the product of the $$\ldots \ldots$$ of the radius and the gradient of the $$\ldots \ldots$$ is equal to $$\ldots \ldots$$.

A circle with centre $$C(a;b)$$ and a radius of $$r$$ units is shown in the diagram above. $$D(x;y)$$ is a point on the circumference and the equation of the circle is:

$(x - a)^{2} + (y - b)^{2} = r^{2}$

A tangent is a straight line that touches the circumference of a circle at only one place.

The tangent line $$AB$$ touches the circle at $$D$$.

The radius of the circle $$CD$$ is perpendicular to the tangent $$AB$$ at the point of contact $$D$$.

\begin{align*} CD & \perp AB \\ \text{and } C\hat{D}A &= C\hat{D}B = \text{90} ° \end{align*}

The product of the gradient of the radius and the gradient of the tangent line is equal to $$-\text{1}$$.

$m_{CD} \times m_{AB} = - 1$

How to determine the equation of a tangent:

1. Determine the equation of the circle and write it in the form $(x - a)^{2} + (y - b)^{2} = r^{2}$
2. From the equation, determine the coordinates of the centre of the circle $$(a;b)$$.
3. Determine the gradient of the radius: $m_{CD} = \frac{y_{2} - y_{1}}{x_{2}- x_{1}}$
4. The radius is perpendicular to the tangent of the circle at a point $$D$$ so: $m_{AB} = - \frac{1}{m_{CD}}$
5. Write down the gradient-point form of a straight line equation and substitute $$m_{AB}$$ and the coordinates of $$D$$. Make $$y$$ the subject of the equation. $y - y_{1} = m(x - x_{1})$

## Worked example 12: Equation of a tangent to a circle

Determine the equation of the tangent to the circle $$x^{2} + y^{2} - 2y + 6x - 7 = 0$$ at the point $$F(-2;5)$$.

### Write the equation of the circle in the form $$(x - a)^{2} + (y - b)^{2} = r^{2}$$

Use the method of completing the square:

\begin{align*} x^{2} + y^{2} - 2y + 6x - 7 &= 0 \\ x^{2} + 6x + y^{2} - 2y &= 7 \\ (x^{2} + 6x + 9) - 9 + (y^{2} - 2y + 1) - 1 &= 7 \\ (x + 3)^{2} + (y - 1)^{2} &= 17 \end{align*}

### Draw a sketch

The centre of the circle is $$(-3;1)$$ and the radius is $$\sqrt{17}$$ units.

### Determine the gradient of the radius $$CF$$

\begin{align*} m_{CF} &= \frac{y_{2} - y_{1}}{x_{2}- x_{1}}\\ &= \frac{5 - 1}{-2 + 3}\\ &= 4 \end{align*}

### Determine the gradient of the tangent

Let the gradient of the tangent line be $$m$$.

\begin{align*} m_{CF} \times m &= -1 \\ 4 \times m &= -1 \\ \therefore m &= - \frac{1}{4} \end{align*}

### Determine the equation of the tangent to the circle

Write down the gradient-point form of a straight line equation and substitute $$m = - \frac{1}{4}$$ and $$F(-2;5)$$.

\begin{align*} y - y_{1} &= m (x - x_{1}) \\ y - y_{1} &= - \frac{1}{4} (x - x_{1}) \\ \text{Substitute } F(-2;5): \quad y - 5 &= - \frac{1}{4} (x - (-2)) \\ y - 5 &= - \frac{1}{4} (x + 2) \\ y &= - \frac{1}{4}x - \frac{1}{2} + 5 \\ &= - \frac{1}{4}x + \frac{9}{2} \end{align*}

The equation of the tangent to the circle at $$F$$ is $$y = - \frac{1}{4}x + \frac{9}{2}$$.

## Worked example 13: Equation of a tangent to a circle

The straight line $$y = x + 4$$ cuts the circle $$x^{2} + y^{2} = 26$$ at $$P$$ and $$Q$$.

1. Calculate the coordinates of $$P$$ and $$Q$$.
2. Sketch the circle and the straight line on the same system of axes. Label points $$P$$ and $$Q$$.
3. Determine the coordinates of $$H$$, the mid-point of chord $$PQ$$.
4. If $$O$$ is the centre of the circle, show that $$PQ \perp OH$$.
5. Determine the equations of the tangents to the circle at $$P$$ and $$Q$$.
6. Determine the coordinates of $$S$$, the point where the two tangents intersect.
7. Show that $$S$$, $$H$$ and $$O$$ are on a straight line.

### Determine the coordinates of $$P$$ and $$Q$$

Substitute the straight line $$y = x + 4$$ into the equation of the circle and solve for $$x$$:

\begin{align*} x^{2} + y^{2} &= 26 \\ x^{2} + (x + 4)^{2} &= 26 \\ x^{2} + x^{2} + 8x + 16 &= 26 \\ 2x^{2} + 8x - 10 &= 0 \\ x^{2} + 4x - 5 &= 0 \\ (x - 1)(x + 5) &= 0 \\ \therefore x = 1 &\text{ or } x = -5 \\ \text{If } x = 1 \quad y &= 1 + 4 = 5 \\ \text{If } x = -5 \quad y &= -5 + 4 = -1 \end{align*}

This gives the points $$P(-5;-1)$$ and $$Q(1;5)$$.

### Determine the coordinates of the mid-point $$H$$

\begin{align*} H(x;y) &= \left( \frac{x_{1} + x_{2}}{2}; \frac{y_{1} + y_{2}}{2} \right) \\ &= \left( \frac{1 - 5}{2}; \frac{5 - 1}{2} \right) \\ &= \left( \frac{-4}{2}; \frac{4}{2} \right) \\ &= \left( -2; 2 \right) \end{align*}

### Show that $$OH$$ is perpendicular to $$PQ$$

We need to show that the product of the two gradients is equal to $$-\text{1}$$. From the given equation of $$PQ$$, we know that $$m_{PQ} = 1$$.

\begin{align*} m_{OH} &= \frac{2 - 0}{-2 - 0} \\ &= - 1 \\ & \\ m_{PQ} \times m_{OH} &= - 1 \\ & \\ \therefore PQ & \perp OH \end{align*}

### Determine the equations of the tangents at $$P$$ and $$Q$$

Tangent at $$P$$:

Determine the gradient of the radius $$OP$$:

\begin{align*} m_{OP} &= \frac{-1 - 0}{- 5 - 0} \\ &= \frac{1}{5} \end{align*}

The tangent of a circle is perpendicular to the radius, therefore we can write:

\begin{align*} \frac{1}{5} \times m_{P} &= -1 \\ \therefore m_{P} &= - 5 \end{align*}

Substitute $$m_{P} = - 5$$ and $$P(-5;-1)$$ into the equation of a straight line.

\begin{align*} y - y_{1} &= - 5 (x - x_{1}) \\ \text{Substitute } P(-5;-1): \quad y + 1 &= - 5 (x + 5) \\ y &= -5x - 25 - 1 \\ &= -5x - 26 \end{align*}

Tangent at $$Q$$:

Determine the gradient of the radius $$OQ$$:

\begin{align*} m_{OQ} &= \frac{5 - 0}{1 - 0} \\ &= 5 \end{align*}

The tangent of a circle is perpendicular to the radius, therefore we can write:

\begin{align*} 5 \times m_{Q} &= -1 \\ \therefore m_{Q} &= - \frac{1}{5} \end{align*}

Substitute $$m_{Q} = - \frac{1}{5}$$ and $$Q(1;5)$$ into the equation of a straight line.

\begin{align*} y - y_{1} &= - \frac{1}{5} (x - x_{1}) \\ \text{Substitute } Q(1;5): \quad y - 5 &= - \frac{1}{5} (x - 1) \\ y &= - \frac{1}{5}x + \frac{1}{5} + 5 \\ &= - \frac{1}{5}x + \frac{26}{5} \end{align*}

The equations of the tangents are $$y = -5x - 26$$ and $$y = - \frac{1}{5}x + \frac{26}{5}$$.

### Determine the coordinates of $$S$$

Equate the two linear equations and solve for $$x$$:

\begin{align*} -5x - 26 &= - \frac{1}{5}x + \frac{26}{5} \\ -25x - 130 &= - x + 26 \\ -24x &= 156 \\ x &= - \frac{156}{24} \\ &= - \frac{13}{2} \\ \text{If } x = - \frac{13}{2} \quad y &= - 5 \left( - \frac{13}{2} \right) - 26 \\ &= \frac{65}{2} - 26 \\ &= \frac{13}{2} \end{align*}

This gives the point $$S \left( - \frac{13}{2}; \frac{13}{2} \right)$$.

### Show that $$S$$, $$H$$ and $$O$$ are on a straight line

We need to show that there is a constant gradient between any two of the three points. We have already shown that $$PQ$$ is perpendicular to $$OH$$, so we expect the gradient of the line through $$S$$, $$H$$ and $$O$$ to be $$-\text{1}$$.

\begin{align*} m_{SH} &= \dfrac{\frac{13}{2} - 2}{- \frac{13}{2} + 2} \\ &= - 1 \end{align*}\begin{align*} m_{SO} &= \dfrac{\frac{13}{2} - 0}{- \frac{13}{2} - 0} \\ &= - 1 \end{align*}

Therefore $$S$$, $$H$$ and $$O$$ all lie on the line $$y=-x$$.

## Worked example 14: Equation of a tangent to a circle

Determine the equations of the tangents to the circle $$x^{2} + (y - 1)^{2} = 80$$, given that both are parallel to the line $$y = \frac{1}{2}x + 1$$.

### Draw a sketch

The tangents to the circle, parallel to the line $$y = \frac{1}{2}x + 1$$, must have a gradient of $$\frac{1}{2}$$. From the sketch we see that there are two possible tangents.

### Determine the coordinates of $$A$$ and $$B$$

To determine the coordinates of $$A$$ and $$B$$, we must find the equation of the line perpendicular to $$y = \frac{1}{2}x + 1$$ and passing through the centre of the circle. This perpendicular line will cut the circle at $$A$$ and $$B$$.

Notice that the line passes through the centre of the circle.

To determine the coordinates of $$A$$ and $$B$$, we substitute the straight line $$y = - 2x + 1$$ into the equation of the circle and solve for $$x$$:

\begin{align*} x^{2} + (y-1)^{2} &= 80 \\ x^{2} + \left( - 2x + 1 - 1 \right)^{2} &= 80 \\ x^{2} + 4x^{2} &= 80 \\ 5x^{2} &= 80 \\ x^{2} &= 16 \\ \therefore x &= \pm 4 \\ \text{If } x = 4 \quad y &= - 2(4) + 1 = - 7 \\ \text{If } x = -4 \quad y &= - 2(-4) + 1 = 9 \end{align*}

This gives the points $$A(-4;9)$$ and $$B(4;-7)$$.

### Determine the equations of the tangents to the circle

Tangent at $$A$$:

\begin{align*} y - y_{1} &= \frac{1}{2} (x - x_{1}) \\ y - 9 &= \frac{1}{2} (x + 4 ) \\ y &= \frac{1}{2} x + 11 \end{align*}

Tangent at $$B$$:

\begin{align*} y - y_{1} &= \frac{1}{2} (x - x_{1}) \\ y + 7 &= \frac{1}{2} (x - 4 ) \\ y &= \frac{1}{2}x - 9 \end{align*}

The equation of the tangent at point $$A$$ is $$y = \frac{1}{2}x + 11$$ and the equation of the tangent at point $$B$$ is $$y = \frac{1}{2}x - 9$$.

## Worked example 15: Equation of a tangent to a circle

Determine the equations of the tangents to the circle $$x^{2} + y^{2} = 25$$, from the point $$G(-7;-1)$$ outside the circle.

### Consider where the two tangents will touch the circle

Let the two tangents from $$G$$ touch the circle at $$F$$ and $$H$$.

\begin{align*} OF = OH &= \text{5}\text{ units} \quad (\text{equal radii}) \\ OG &= \sqrt{(0 + 7)^{2} + (0 + 1)^2} \\ &= \sqrt{50} \\ GF &= \sqrt{ (x + 7)^{2} + (y + 1)^2} \\ \therefore GF^{2} &= (x + 7)^{2} + (y + 1)^2 \\ \text{And } G\hat{F}O = G\hat{H}O &= \text{90} ° \end{align*}

Consider $$\triangle GFO$$ and apply the theorem of Pythagoras:

\begin{align*} GF^{2} + OF^{2} &= OG^{2} \\ \left( x + 7 \right)^{2} + \left( y + 1 \right)^{2} + 5^{2} &= \left( \sqrt{50} \right)^{2} \\ x^{2} + 14x + 49 + y^{2} + 2y + 1 + 25 &= 50 \\ x^{2} + 14x + y^{2} + 2y + 25 &= 0 \ldots \ldots (1) \\ \text{Substitute } y^{2} = 25 - x^{2} & \text{ into equation } (1) \\ \quad x^{2} + 14x + \left( 25 - x^{2} \right) + 2\left( \sqrt{25 - x^{2}} \right) + 25 &= 0 \\ 14x + 50 &= - 2\left( \sqrt{25 - x^{2}} \right) \\ 7x + 25 &= - \sqrt{25 - x^{2}} \\ \text{Square both sides: } (7x + 25)^{2} &= \left( - \sqrt{25 - x^{2}} \right)^{2} \\ 49x^{2} + 350x + 625 &= 25 - x^{2} \\ 50x^{2} + 350x + 600 &= 0 \\ x^{2} + 7x + 12 &= 0 \\ (x + 3)(x + 4) &= 0 \\ \therefore x = -3 & \text{ or } x = -4 \\ \text{At } F: x = -3 \quad y &= - \sqrt{25 - (-3)^{2}} = - \sqrt{16} = - 4 \\ \text{At } H: x = -4 \quad y &= \sqrt{25 - (-4)^{2}} = \sqrt{9} = 3 \end{align*}

Note: from the sketch we see that $$F$$ must have a negative $$y$$-coordinate, therefore we take the negative of the square root. Similarly, $$H$$ must have a positive $$y$$-coordinate, therefore we take the positive of the square root.

This gives the points $$F(-3;-4)$$ and $$H(-4;3)$$.

Tangent at $$F$$:

\begin{align*} m_{FG} &= \frac{-1 + 4}{-7 + 3} \\ &= - \frac{3}{4} \end{align*}\begin{align*} y - y_{1} &= m (x - x_{1}) \\ y - y_{1} &= - \frac{3}{4} (x - x_{1}) \\ y + 1 &= - \frac{3}{4} (x + 7) \\ y &= - \frac{3}{4}x - \frac{21}{4} - 1 \\ y &= - \frac{3}{4}x - \frac{25}{4} \end{align*}

Tangent at $$H$$:

\begin{align*} m_{HG} &= \frac{-1 - 3}{-7 + 4} \\ &= \frac{4}{3} \end{align*}\begin{align*} y + 1 &= \frac{4}{3} \left(x + 7 \right) \\ y &= \frac{4}{3}x + \frac{28}{3} - 1 \\ y &= \frac{4}{3}x + \frac{25}{3} \end{align*}

The equations of the tangents to the circle are $$y = - \frac{3}{4}x - \frac{25}{4}$$ and $$y = \frac{4}{3}x + \frac{25}{3}$$.

## Equation of a tangent to a circle

Exercise 7.5

A circle with centre $$(8;-7)$$ and the point $$(5;-5)$$ on the circle are given. Determine the gradient of the radius.

Given:

• the centre of the circle $$(a;b) = (8;-7)$$
• a point on the circumference of the circle $$(x_1;y_1) = (5;-5)$$

Required:

• the gradient of the radius, $$m$$
\begin{align*} m & = \frac{y_2 - y_1}{x_2 - x_1}\\ & = \frac{-5+7}{5-8}\\ & = - \frac{2}{3} \end{align*}

The gradient of the radius is $$m = - \frac{2}{3}$$.

Determine the gradient of the tangent to the circle at the point $$(5;-5)$$.

The tangent to the circle at the point $$(5;-5)$$ is perpendicular to the radius of the circle to that same point: $$m \times m_{\bot} = -1$$

\begin{align*} m_{\bot} & = - \frac{1}{m}\\ & = \frac{-1}{- \frac{2}{3} }\\ & = \frac{3}{2} \end{align*}

The gradient for the tangent is $$m_{\bot} = \frac{3}{2}$$.

Given the equation of the circle: $$\left(x + 4\right)^{2} + \left(y + 8\right)^{2} = 136$$

Find the gradient of the radius at the point $$(2;2)$$ on the circle.

Given:

• the equation for the circle $$\left(x + 4\right)^{2} + \left(y + 8\right)^{2} = 136$$
• a point on the circumference of the circle $$(x_1;y_1) = (2;2)$$

Required:

• the gradient of the radius, $$m$$

The coordinates of the centre of the circle are $$(-4;-8)$$.

Draw a rough sketch:

The gradient for this radius is $$m = \frac{5}{3}$$.

Determine the gradient of the tangent to the circle at the point $$(2;2)$$.

Given:

The tangent to the circle at the point $$(2;2)$$ is perpendicular to the radius, so $$m \times m_{\text{tangent}} = -1$$

\begin{align*} m_{\text{tangent}} & = - \frac{1}{m}\\ & = - \frac{1}{\frac{5}{3} }\\ & = - \frac{3}{5} \end{align*}

The gradient for the tangent is $$m_{\text{tangent}} = - \frac{3}{5}$$.

Given a circle with the central coordinates $$(a;b) = (-9;6)$$. Determine the equation of the tangent to the circle at the point $$(-2;5)$$.

\begin{align*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{5 - 6 }{ -2 -(-9)} \\ & = - \frac{1}{7} \end{align*}

The tangent is perpendicular to the radius, therefore $$m \times m_{\bot} = -1$$.

\begin{align*} m & = - \frac{1}{m_r} \\ & = \frac{1}{ \frac{1}{7} } \\ & = 7 \end{align*}

Write down the equation of a straight line and substitute $$m = 7$$ and $$(-2;5)$$.

\begin{align*} y_1 & =m x_1 + c\\ 5 & = 7 (-2) + c \\ c & = 19 \end{align*}

The equation of the tangent to the circle is $$y = 7 x + 19$$.

Given the diagram below:

Determine the equation of the tangent to the circle with centre $$C$$ at point $$H$$.

Given:

• the centre of the circle $$C(a;b) = (1;5)$$
• a point on the circumference of the circle $$H(-2;1)$$

Required:

• the equation for the tangent to the circle in the form $$y = mx + c$$

\begin{align*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{1 - 5}{-2 - 1 } \\ & = \frac{-4}{-3 } \\ & = \frac{4}{3} \end{align*} \begin{align*} m_r \times m &= -1 \\ m & = - \frac{1}{m_r} \\ & = - \frac{1}{\frac{4}{3} } \\ & = - \frac{3}{4} \end{align*}

Equation of the tangent:

\begin{align*} y & = m x + c\\ 1 & = - \frac{3}{4} (-2) + c \\ 1 & = \frac{3}{2} + c \\ c & = - \frac{1}{2} \end{align*}

The equation for the tangent to the circle at the point $$H$$ is:

\begin{align*} y & = - \frac{3}{4} x - \frac{1}{2} \end{align*}

Given the point $$P(2;-4)$$ on the circle $$\left(x - 4\right)^{2} + \left(y + 5\right)^{2} = 5$$. Find the equation of the tangent at $$P$$.

Given:

• the equation for the circle $$\left(x - 4\right)^{2} + \left(y + 5\right)^{2} = 5$$
• a point on the circumference of the circle $$P(2;-4)$$

Required:

• the equation of the tangent in the form $$y = mx + c$$

The coordinates of the centre of the circle are $$(a;b) = (4;-5)$$.

\begin{align*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{ -4 - (-5)}{2 - 4} \\ & = - \frac{1}{2} \end{align*} \begin{align*} m \times m_{\bot} & = -1 \\ \therefore m_{\bot} & = - \frac{1}{m_r} \\ & = \frac{1}{\frac{1}{2}} \\ & = 2 \end{align*}

Equation of the tangent:

\begin{align*} y & = m_{\bot} x + c\\ -4 & = 2 (2) + c \\ c & = -8 \end{align*}

The equation of the tangent to the circle is

\begin{align*} y & = 2 x - 8 \end{align*}

$$C(-4;8)$$ is the centre of the circle passing through $$H(2;-2)$$ and $$Q(-10;m)$$.

Determine the equation of the circle.

Use the distance formula to determine the length of the radius:

\begin{align*} r & = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ & = \sqrt{(2+4)^2 + (-2-8)^2} \\ & = \sqrt{(6)^2 + (-10)^2} \\ & = \sqrt{136} \end{align*}

Write down the general equation of a circle and substitute $$r$$ and $$H(2;-2)$$:

\begin{align*} (x-a)^2+ (y-b)^2 & = r^2 \\ (x - (-4))^2 + (y-(8))^2 & = (\sqrt{136})^2 \\ \left(x + 4\right)^{2} + \left(y - 8\right)^{2} & = 136 \end{align*}

The equation of the circle is $$\left(x + 4\right)^{2} + \left(y - 8\right)^{2} = 136$$.

Determine the value of $$m$$.

Substitute the $$Q(-10;m)$$ and solve for the $$m$$ value.

\begin{align*} \left(x + 4\right)^{2} + \left(y - 8\right)^{2} & = 136 \\ \left(-10 + 4\right)^{2} + \left(m - 8\right)^{2} & = 136 \\ 36 + \left(m - 8\right)^{2} & = 136 \\ m^{2} - 16 m + 100 & = 136 \\ m^{2} - 16 m - 36 & = 0 \\ (m+2)(m-18) & = 0 \end{align*}

The solution shows that $$y = -2$$ or $$y = 18$$. From the graph we see that the $$y$$-coordinate of $$Q$$ must be positive, therefore $$Q(-10;18)$$.

Determine the equation of the tangent to the circle at point $$Q$$.

\begin{align*} m_r & = \frac{y_2 - y_0}{x_2 - x_0} \\ & = \frac{18 - 8}{-10 + 4 } \\ & = - \frac{10}{6 } \\ & = - \frac{5}{3} \end{align*}

The radius is perpendicular to the tangent, so $$m \times m_{\bot} = -1$$.

\begin{align*} m_{\bot} & = - \frac{1}{m_r} \\ & = \frac{1}{\frac{5}{3}} \\ & = \frac{3}{5} \end{align*}

The equation for the tangent to the circle at the point $$Q$$ is:

\begin{align*} y & = m_{\bot} x + c\\ 18 & = \frac{3}{5} (-10) + c \\ 18 & = -6 + c \\ c & = 24 \end{align*} \begin{align*} y & = \frac{3}{5} x + 24 \end{align*}

The straight line $$y = x + 2$$ cuts the circle $$x^{2} + y^{2} = 20$$ at $$P$$ and $$Q$$.

Calculate the coordinates of $$P$$ and $$Q$$.

Substitute the straight line $$y = x + 2$$ into the equation of the circle and solve for $$x$$:

\begin{align*} x^{2} + y^{2} &= 20 \\ x^{2} + (x + 2)^{2} &= 20 \\ x^{2} + x^{2} + 4x + 4 &= 20 \\ 2x^{2} + 4x - 16 &= 0 \\ x^{2} + 2x - 8 &= 0 \\ (x - 2)(x + 4) &= 0 \\ \therefore x = 2 &\text{ or } x = -4 \\ \text{If } x = 2 \quad y &= 2 + 2 = 4 \\ \text{If } x = -4 \quad y &= -4 + 2 = -2 \end{align*}

This gives the points $$P(-4;-2)$$ and $$Q(2;4)$$.

Determine the length of $$PQ$$.
\begin{align*} PQ &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ &= \sqrt{(-4 -2)^{2} + (-2-4 )^2} \\ &= \sqrt{(-6)^{2} + (-6)^2} \\ &= \sqrt{36 + 36} \\ &= \sqrt{36 \cdot 2} \\ &= 6\sqrt{2} \end{align*}
Determine the coordinates of $$M$$, the mid-point of chord $$PQ$$.
\begin{align*} M(x;y) &= \left( \frac{x_{1} + x_{2}}{2}; \frac{y_{1} + y_{2}}{2} \right) \\ &= \left( \frac{-4 + 2}{2}; \frac{-2 + 4}{2} \right) \\ &= \left( \frac{-2}{2}; \frac{2}{2} \right) \\ &= \left( -1; 1 \right) \end{align*}
If $$O$$ is the centre of the circle, show that $$PQ \perp OM$$.
\begin{align*} m_{PQ} &= \frac{4 - (-2)}{2 - (-4)} \\ &= \frac{6}{6} \\ &= 1 \\ & \\ m_{OM} &= \frac{1 - 0}{-1 - 0} \\ &= - 1 \\ m_{PQ} \times m_{OM} &= - 1 \\ & \\ \therefore PQ & \perp OM \end{align*}
Determine the equations of the tangents to the circle at $$P$$ and $$Q$$.

Tangent at $$P$$:

Determine the gradient of the radius $$OP$$:

\begin{align*} m_{OP} &= \frac{y_{2} - y_{1}}{x_{2}- x_{1}} \\ &= \frac{-2 - 0}{- 4 - 0} \\ &= \frac{1}{2} \end{align*}

Let the gradient of the tangent at $$P$$ be $$m_{P}$$. The tangent of a circle is perpendicular to the radius, therefore we can write:

\begin{align*} m_{OP} \times m_{P} &= -1 \\ \frac{1}{2} \times m_{P} &= -1 \\ \therefore m_{P} &= - 2 \end{align*}

Substitute $$m_{P} = - 2$$ and $$P(-4;-2)$$ into the equation of a straight line.

\begin{align*} y - y_{1} &= m (x - x_{1}) \\ y - y_{1} &= - 2 (x - x_{1}) \\ \text{Substitute } P(-4;-2): \quad y + 2 &= - 2 (x + 4) \\ y &= -2x - 8 - 2 \\ &= -2x - 10 \end{align*}

Tangent at $$Q$$:

Determine the gradient of the radius $$OQ$$:

\begin{align*} m_{OQ} &= \frac{y_{2} - y_{1}}{x_{2}- x_{1}} \\ &= \frac{4 - 0}{2 - 0} \\ &= 2 \end{align*}

Let the gradient of the tangent at $$Q$$ be $$m_{Q}$$. The tangent of a circle is perpendicular to the radius, therefore we can write:

\begin{align*} m_{OQ} \times m_{Q} &= -1 \\ 2 \times m_{Q} &= -1 \\ \therefore m_{Q} &= - \frac{1}{2} \end{align*}

Substitute $$m_{Q} = - \frac{1}{2}$$ and $$Q(2;4)$$ into the equation of a straight line.

\begin{align*} y - y_{1} &= m (x - x_{1}) \\ y - y_{1} &= - \frac{1}{2} (x - x_{1}) \\ \text{Substitute } Q(2;4): \quad y - 4 &= - \frac{1}{2} (x - 2) \\ y &= - \frac{1}{2}x + 1 + 4 \\ &= - \frac{1}{2}x + 5 \end{align*}

Therefore the equations of the tangents to the circle are $$y = -2x - 10$$ and $$y = - \frac{1}{2}x + 5$$.

Determine the coordinates of $$S$$, the point where the two tangents intersect.

Equate the two linear equations and solve for $$x$$:

\begin{align*} -2x - 10 &= - \frac{1}{2}x + 5 \\ -4x - 20 &= - x + 10 \\ -3x &= 30 \\ x &= - 10 \\ \text{If } x = - 10 \quad y &= - 2 \left( - 10 \right) - 10 \\ &= 10 \end{align*}

This gives the point $$S \left( - 10;10 \right)$$.

Show that $$PS = QS$$.
\begin{align*} PS &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ &= \sqrt{(-4 -(-10))^{2} + (-2 - 10)^2} \\ &= \sqrt{(6)^{2} + (-12)^2} \\ &= \sqrt{36 + 144} \\ &= \sqrt{180} \end{align*} \begin{align*} QS &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ &= \sqrt{(2 -(-10))^{2} + (4 - 10)^2} \\ &= \sqrt{(12)^{2} + (-6)^2} \\ &= \sqrt{144 + 36} \\ &= \sqrt{180} \end{align*}
Determine the equations of the two tangents to the circle, both parallel to the line $$y + 2x = 4$$.

The tangent at $$P$$, $$y = -2x - 10$$, is parallel to $$y = - 2x + 4$$. To find the equation of the second parallel tangent:

\begin{align*} y &= -2x + 4 \\ \therefore m &= -2 \\ \therefore m_{\text{radius}}&= \frac{1}{2} \\ \text{Eqn. of radius: } y &= \frac{1}{2}x \ldots(1) \\ \text{Substitute } (1): \quad x^{2} + y^{2} &= 20 \\ x^{2} + \left( \frac{1}{2}x \right)^{2} &= 20 \\ x^{2} + \frac{1}{4}x^{2} &= 20 \\ \frac{5}{4}x^{2} &= 20 \\ x^{2} &= 16 \\ x &= \pm 4 \\ \text{If } x = 4, y &= 2 \\ \text{Substitute } (4;2): \quad y &= -2x + c \\ 2 &=-2(4) + c \\ 10 &= c \\ y &= -2x + 10 \end{align*}