mc009-2.jpg is tangent to circle A at B and to circle D at C. What is AD to the nearest tenth?

seven.iii Equation of a tangent to a circumvolve (EMCHW)

1. On a suitable system of axes, draw the circle \(x^{ii} + y^{two} = twenty\) with centre at \(O(0;0)\).
2. Plot the point \(T(2;4)\).
3. Plot the bespeak \(P(0;five)\). Draw \(PT\) and extend the line and so that is cuts the positive \(10\)-axis.
4. Measure out \(O\hat{T}P\).
5. Determine the gradient of the radius \(OT\).
6. Determine the slope of \(PT\).
7. Evidence that \(PT \perp OT\).
8. Plot the betoken \(Southward(2;-4)\) and join \(OS\).
9. Draw a tangent to the circle at \(Southward\).
10. Measure the bending between \(Os\) and the tangent line at \(Due south\).
11. Make a conjecture almost the bending betwixt the radius and the tangent to a circle at a point on the circumvolve.
12. Complete the judgement: 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:

\[(x - a)^{2} + (y - b)^{ii} = r^{ii}\]

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

The tangent line \(AB\) touches the circumvolve at \(D\).

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

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

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

\[m_{CD} \times m_{AB} = - one\]

How to decide the equation of a tangent:

1. Determine the equation of the circumvolve and write it in the form \[(x - a)^{2} + (y - b)^{ii} = r^{2}\]
2. From the equation, determine the coordinates of the middle of the circle \((a;b)\).
3. Decide 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 downwardly the gradient-point class of a straight line equation and substitute \(m_{AB}\) and the coordinates of \(D\). Brand \(y\) the bailiwick of the equation. \[y - y_{1} = m(x - x_{one})\]

Worked example 12: Equation of a tangent to a circumvolve

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

Write the equation of the circle in the grade \((10 - a)^{2} + (y - b)^{2} = r^{2}\)

Use the method of completing the square:

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

Draw a sketch

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

Determine the gradient of the radius \(CF\)

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

Determine the gradient of the tangent

Allow the slope of the tangent line be \(m\).

\begin{align*} m_{CF} \times m &= -i \\ four \times m &= -1 \\ \therefore grand &= - \frac{ane}{4} \terminate{align*}

Make up one's mind the equation of the tangent to the circle

Write down the gradient-point course of a straight line equation and substitute \(one thousand = - \frac{one}{4}\) and \(F(-2;5)\).

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

Write the last answer

The equation of the tangent to the circle at \(F\) is \(y = - \frac{1}{4}x + \frac{ix}{ii}\).

temp text

Worked example 13: Equation of a tangent to a circle

The straight line \(y = x + four\) cuts the circle \(10^{ii} + y^{2} = 26\) at \(P\) and \(Q\).

1. Summate the coordinates of \(P\) and \(Q\).
2. Sketch the circumvolve and the straight line on the same system of axes. Label points \(P\) and \(Q\).
3. Decide the coordinates of \(H\), the mid-indicate 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 directly line.

Determine the coordinates of \(P\) and \(Q\)

Substitute the straight line \(y = x + 4\) into the equation of the circumvolve and solve for \(10\):

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

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

Draw a sketch

Determine the coordinates of the mid-indicate \(H\)

\begin{align*} H(x;y) &= \left( \frac{x_{1} + x_{2}}{2}; \frac{y_{i} + y_{2}}{ii} \correct) \\ &= \left( \frac{ane - 5}{2}; \frac{5 - 1}{ii} \correct) \\ &= \left( \frac{-iv}{2}; \frac{4}{2} \right) \\ &= \left( -2; ii \correct) \cease{align*}

Show that \(OH\) is perpendicular to \(PQ\)

We demand to prove that the product of the two gradients is equal to \(-\text{1}\). From the given equation of \(PQ\), we know that \(m_{PQ} = 1\).

\brainstorm{marshal*} m_{OH} &= \frac{2 - 0}{-2 - 0} \\ &= - i \\ & \\ m_{PQ} \times m_{OH} &= - 1 \\ & \\ \therefore PQ & \perp OH \finish{align*}

Decide the equations of the tangents at \(P\) and \(Q\)

Tangent at \(P\):

Make up one's mind the gradient of the radius \(OP\):

\brainstorm{align*} m_{OP} &= \frac{-i - 0}{- 5 - 0} \\ &= \frac{1}{5} \stop{align*}

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

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

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

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

Tangent at \(Q\):

Make up one's mind the gradient of the radius \(OQ\):

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

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

\begin{align*} 5 \times m_{Q} &= -ane \\ \therefore m_{Q} &= - \frac{one}{5} \finish{align*}

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

\begin{align*} y - y_{1} &= - \frac{1}{5} (x - x_{1}) \\ \text{Substitute } Q(1;v): \quad y - 5 &= - \frac{one}{v} (x - 1) \\ y &= - \frac{1}{5}10 + \frac{1}{5} + 5 \\ &= - \frac{1}{five}x + \frac{26}{v} \end{align*}

The equations of the tangents are \(y = -5x - 26\) and \(y = - \frac{1}{5}10 + \frac{26}{v}\).

Decide the coordinates of \(South\)

Equate the two linear equations and solve for \(x\):

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

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

Show that \(Due south\), \(H\) 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 look the gradient of the line through \(S\), \(H\) and \(O\) to be \(-\text{1}\).

\brainstorm{align*} m_{SH} &= \dfrac{\frac{13}{two} - 2}{- \frac{thirteen}{ii} + ii} \\ &= - 1 \end{align*}\brainstorm{align*} m_{SO} &= \dfrac{\frac{13}{2} - 0}{- \frac{13}{2} - 0} \\ &= - 1 \end{align*}

Therefore \(South\), \(H\) and \(O\) all lie on the line \(y=-x\).

Worked example xiv: Equation of a tangent to a circumvolve

Determine the equations of the tangents to the circle \(10^{2} + (y - 1)^{2} = fourscore\), given that both are parallel to the line \(y = \frac{1}{2}10 + ane\).

Depict a sketch

The tangents to the circumvolve, parallel to the line \(y = \frac{ane}{ii}x + 1\), must have a slope of \(\frac{one}{2}\). From the sketch nosotros run into that at that place are ii possible tangents.

Make up one's mind the coordinates of \(A\) and \(B\)

To determine the coordinates of \(A\) and \(B\), we must detect the equation of the line perpendicular to \(y = \frac{ane}{2}ten + 1\) and passing through the center of the circle. This perpendicular line will cut the circle at \(A\) and \(B\).

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

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

\brainstorm{align*} x^{two} + (y-1)^{ii} &= 80 \\ x^{ii} + \left( - 2x + 1 - ane \right)^{ii} &= fourscore \\ ten^{two} + 4x^{2} &= 80 \\ 5x^{ii} &= 80 \\ ten^{2} &= 16 \\ \therefore 10 &= \pm 4 \\ \text{If } ten = 4 \quad y &= - 2(4) + 1 = - 7 \\ \text{If } x = -4 \quad y &= - 2(-4) + one = ix \finish{marshal*}

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

Determine the equations of the tangents to the circumvolve

Tangent at \(A\):

\begin{align*} y - y_{1} &= \frac{one}{two} (x - x_{i}) \\ y - 9 &= \frac{ane}{2} (x + iv ) \\ y &= \frac{1}{two} x + 11 \end{align*}

Tangent at \(B\):

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

The equation of the tangent at indicate \(A\) is \(y = \frac{1}{2}x + xi\) and the equation of the tangent at point \(B\) is \(y = \frac{1}{2}x - ix\).

Worked example fifteen: Equation of a tangent to a circumvolve

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

Depict a sketch

Consider where the two tangents will touch the circle

Allow the two tangents from \(G\) affect the circle at \(F\) and \(H\).

\begin{align*} OF = OH &= \text{five}\text{ units} \quad (\text{equal radii}) \\ OG &= \sqrt{(0 + 7)^{2} + (0 + i)^2} \\ &= \sqrt{l} \\ GF &= \sqrt{ (x + 7)^{ii} + (y + 1)^2} \\ \therefore GF^{2} &= (ten + 7)^{ii} + (y + 1)^ii \\ \text{And } 1000\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)^{two} + \left( y + 1 \correct)^{2} + 5^{ii} &= \left( \sqrt{fifty} \right)^{2} \\ x^{2} + 14x + 49 + y^{2} + 2y + ane + 25 &= fifty \\ ten^{2} + 14x + y^{2} + 2y + 25 &= 0 \ldots \ldots (ane) \\ \text{Substitute } y^{2} = 25 - ten^{two} & \text{ into equation } (i) \\ \quad x^{2} + 14x + \left( 25 - x^{2} \right) + 2\left( \sqrt{25 - 10^{two}} \right) + 25 &= 0 \\ 14x + fifty &= - 2\left( \sqrt{25 - x^{2}} \right) \\ 7x + 25 &= - \sqrt{25 - ten^{2}} \\ \text{Square both sides: } (7x + 25)^{2} &= \left( - \sqrt{25 - ten^{ii}} \right)^{two} \\ 49x^{2} + 350x + 625 &= 25 - x^{2} \\ 50x^{ii} + 350x + 600 &= 0 \\ x^{two} + 7x + 12 &= 0 \\ (10 + 3)(ten + four) &= 0 \\ \therefore x = -3 & \text{ or } x = -iv \\ \text{At } F: x = -3 \quad y &= - \sqrt{25 - (-3)^{2}} = - \sqrt{16} = - four \\ \text{At } H: x = -4 \quad y &= \sqrt{25 - (-4)^{2}} = \sqrt{9} = three \end{align*}

Notation: from the sketch we run into that \(F\) must have a negative \(y\)-coordinate, therefore we take the negative of the foursquare root. Similarly, \(H\) must have a positive \(y\)-coordinate, therefore we take the positive of the square root.

This gives the points \(F(-3;-four)\) and \(H(-four;3)\).

Tangent at \(F\):

\begin{align*} m_{FG} &= \frac{-1 + four}{-vii + 3} \\ &= - \frac{3}{iv} \end{align*}\brainstorm{align*} y - y_{one} &= m (x - x_{1}) \\ y - y_{1} &= - \frac{three}{4} (x - x_{i}) \\ y + 1 &= - \frac{3}{4} (x + 7) \\ y &= - \frac{3}{4}10 - \frac{21}{iv} - 1 \\ y &= - \frac{iii}{4}x - \frac{25}{four} \end{marshal*}

Tangent at \(H\):

\begin{align*} m_{HG} &= \frac{-one - 3}{-7 + 4} \\ &= \frac{4}{3} \finish{marshal*}\begin{marshal*} y + 1 &= \frac{iv}{iii} \left(10 + 7 \right) \\ y &= \frac{4}{3}x + \frac{28}{3} - ane \\ y &= \frac{4}{3}x + \frac{25}{3} \terminate{align*}

Write the final respond

The equations of the tangents to the circumvolve are \(y = - \frac{3}{4}x - \frac{25}{iv}\) and \(y = \frac{four}{3}x + \frac{25}{3}\).

Equation of a tangent to a circumvolve

Textbook Exercise 7.5

A circumvolve with eye \((8;-7)\) and the point \((5;-5)\) on the circle are given. Decide the slope of the radius.

Given:

• the center of the circle \((a;b) = (8;-seven)\)
• a signal on the circumference of the circle \((x_1;y_1) = (5;-five)\)

Required:

• the slope of the radius, \(grand\)

\begin{align*} m & = \frac{y_2 - y_1}{x_2 - x_1}\\ & = \frac{-5+seven}{5-8}\\ & = - \frac{2}{3} \cease{align*}

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

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

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

\begin{align*} m_{\bot} & = - \frac{1}{one thousand}\\ & = \frac{-ane}{- \frac{2}{3} }\\ & = \frac{three}{2} \end{align*}

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

Discover the slope of the radius at the point \((2;two)\) on the circle.

Given:

• the equation for the circle \(\left(10 + 4\right)^{2} + \left(y + viii\correct)^{two} = 136\)
• a signal on the circumference of the circumvolve \((x_1;y_1) = (two;ii)\)

Required:

• the gradient of the radius, \(m\)

The coordinates of the center of the circle are \((-4;-viii)\).

Draw a crude sketch:

The gradient for this radius is \(yard = \frac{5}{3}\).

Determine the gradient of the tangent to the circle at the betoken \((ii;2)\).

Given:

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

\brainstorm{marshal*} m_{\text{tangent}} & = - \frac{one}{m}\\ & = - \frac{one}{\frac{five}{three} }\\ & = - \frac{three}{5} \cease{marshal*}

The slope for the tangent is \(m_{\text{tangent}} = - \frac{3}{5}\).

Given a circle with the key coordinates \((a;b) = (-9;6)\). Decide 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 - vi }{ -two -(-9)} \\ & = - \frac{ane}{7} \end{align*}

The tangent is perpendicular to the radius, therefore \(1000 \times m_{\bot} = -1\).

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

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

\brainstorm{marshal*} y_1 & =g x_1 + c\\ 5 & = 7 (-2) + c \\ c & = xix \end{align*}

The equation of the tangent to the circle is \(y = 7 x + 19\).

Given the diagram below:

Make up one's mind the equation of the tangent to the circle with centre \(C\) at betoken \(H\).

Given:

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

Required:

• the equation for the tangent to the circle in the form \(y = mx + c\)

Calculate the gradient of the radius:

\brainstorm{align*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{one - 5}{-2 - ane } \\ & = \frac{-four}{-3 } \\ & = \frac{4}{iii} \end{align*} \brainstorm{align*} m_r \times m &= -i \\ thou & = - \frac{1}{m_r} \\ & = - \frac{1}{\frac{4}{3} } \\ & = - \frac{iii}{four} \end{marshal*}

Equation of the tangent:

\begin{align*} y & = g x + c\\ 1 & = - \frac{three}{4} (-ii) + 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} 10 - \frac{1}{2} \stop{marshal*}

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

Given:

• the equation for the circumvolve \(\left(10 - 4\right)^{2} + \left(y + v\correct)^{two} = 5\)
• a point on the circumference of the circle \(P(2;-4)\)

Required:

• the equation of the tangent in the grade \(y = mx + c\)

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

The gradient of the radius:

\begin{align*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{ -4 - (-5)}{2 - four} \\ & = - \frac{1}{2} \stop{align*} \begin{align*} 1000 \times m_{\bot} & = -1 \\ \therefore m_{\bot} & = - \frac{1}{m_r} \\ & = \frac{i}{\frac{i}{2}} \\ & = 2 \end{marshal*}

Equation of the tangent:

\brainstorm{align*} y & = m_{\bot} ten + c\\ -iv & = 2 (2) + c \\ c & = -8 \end{align*}

The equation of the tangent to the circle is

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

Determine the equation of the circumvolve.

Use the distance formula to make up one's mind the length of the radius:

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

Write downwards the full general equation of a circumvolve and substitute \(r\) and \(H(2;-2)\):

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

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

Make up one's mind the value of \(m\).

Substitute the \(Q(-10;thousand)\) and solve for the \(g\) value.

\begin{align*} \left(ten + 4\right)^{2} + \left(y - 8\right)^{2} & = 136 \\ \left(-10 + 4\right)^{two} + \left(m - viii\right)^{2} & = 136 \\ 36 + \left(grand - eight\correct)^{2} & = 136 \\ 1000^{2} - 16 m + 100 & = 136 \\ thousand^{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 run across that the \(y\)-coordinate of \(Q\) must be positive, therefore \(Q(-10;18)\).

Determine the equation of the tangent to the circumvolve at betoken \(Q\).

Calculate the gradient of the radius:

\begin{marshal*} m_r & = \frac{y_2 - y_0}{x_2 - x_0} \\ & = \frac{eighteen - 8}{-10 + 4 } \\ & = - \frac{10}{half dozen } \\ & = - \frac{5}{iii} \cease{align*}

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

\brainstorm{align*} m_{\bot} & = - \frac{1}{m_r} \\ & = \frac{1}{\frac{5}{3}} \\ & = \frac{three}{five} \end{marshal*}

The equation for the tangent to the circumvolve at the bespeak \(Q\) is:

\begin{align*} y & = m_{\bot} x + c\\ 18 & = \frac{3}{v} (-x) + c \\ 18 & = -6 + c \\ c & = 24 \cease{marshal*} \begin{align*} y & = \frac{3}{5} x + 24 \finish{align*}

Calculate the coordinates of \(P\) and \(Q\).

Substitute the direct line \(y = x + 2\) into the equation of the circle and solve for \(10\):

\begin{marshal*} x^{2} + y^{2} &= xx \\ ten^{2} + (x + 2)^{2} &= 20 \\ x^{2} + x^{two} + 4x + four &= 20 \\ 2x^{2} + 4x - 16 &= 0 \\ x^{2} + 2x - 8 &= 0 \\ (x - 2)(x + 4) &= 0 \\ \therefore ten = ii &\text{ or } x = -4 \\ \text{If } x = two \quad y &= ii + 2 = 4 \\ \text{If } x = -4 \quad y &= -4 + 2 = -ii \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})^two} \\ &= \sqrt{(-4 -ii)^{2} + (-2-4 )^2} \\ &= \sqrt{(-vi)^{two} + (-6)^2} \\ &= \sqrt{36 + 36} \\ &= \sqrt{36 \cdot ii} \\ &= vi\sqrt{2} \finish{align*}

Determine the coordinates of \(K\), the mid-signal of chord \(PQ\).

\begin{align*} M(x;y) &= \left( \frac{x_{1} + x_{2}}{2}; \frac{y_{i} + y_{2}}{2} \right) \\ &= \left( \frac{-iv + 2}{2}; \frac{-2 + 4}{2} \correct) \\ &= \left( \frac{-2}{2}; \frac{2}{ii} \right) \\ &= \left( -i; ane \right) \end{align*}

If \(O\) is the eye of the circle, bear witness that \(PQ \perp OM\).

\begin{align*} m_{PQ} &= \frac{4 - (-two)}{two - (-4)} \\ &= \frac{6}{6} \\ &= 1 \\ & \\ m_{OM} &= \frac{one - 0}{-1 - 0} \\ &= - 1 \\ m_{PQ} \times m_{OM} &= - 1 \\ & \\ \therefore PQ & \perp OM \end{align*}

Decide 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_{one}} \\ &= \frac{-2 - 0}{- 4 - 0} \\ &= \frac{1}{2} \cease{align*}

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

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

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

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

Tangent at \(Q\):

Determine the gradient of the radius \(OQ\):

\brainstorm{align*} m_{OQ} &= \frac{y_{2} - y_{1}}{x_{2}- x_{1}} \\ &= \frac{4 - 0}{2 - 0} \\ &= ii \terminate{marshal*}

Permit the gradient of the tangent at \(Q\) exist \(m_{Q}\). The tangent of a circle is perpendicular to the radius, therefore nosotros can write:

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

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

\begin{marshal*} y - y_{ane} &= g (x - x_{ane}) \\ y - y_{1} &= - \frac{i}{2} (x - x_{i}) \\ \text{Substitute } Q(two;4): \quad y - 4 &= - \frac{1}{2} (x - ii) \\ y &= - \frac{ane}{2}ten + ane + 4 \\ &= - \frac{1}{two}ten + 5 \end{align*}

Therefore the equations of the tangents to the circle are \(y = -2x - 10\) and \(y = - \frac{1}{two}ten + v\).

Determine the coordinates of \(S\), the signal where the two tangents intersect.

Equate the ii linear equations and solve for \(x\):

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

This gives the betoken \(S \left( - ten;10 \right)\).

Show that \(PS = QS\).

\begin{marshal*} PS &= \sqrt{(x_{two} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ &= \sqrt{(-4 -(-10))^{2} + (-2 - ten)^2} \\ &= \sqrt{(6)^{2} + (-12)^ii} \\ &= \sqrt{36 + 144} \\ &= \sqrt{180} \stop{align*} \brainstorm{align*} QS &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{i})^2} \\ &= \sqrt{(2 -(-10))^{2} + (4 - 10)^two} \\ &= \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 detect the equation of the second parallel tangent:

\begin{marshal*} y &= -2x + 4 \\ \therefore m &= -2 \\ \therefore m_{\text{radius}}&= \frac{ane}{2} \\ \text{Eqn. of radius: } y &= \frac{1}{2}x \ldots(1) \\ \text{Substitute } (1): \quad x^{2} + y^{two} &= 20 \\ x^{2} + \left( \frac{ane}{2}x \correct)^{two} &= 20 \\ ten^{2} + \frac{1}{4}x^{two} &= 20 \\ \frac{v}{four}ten^{2} &= 20 \\ x^{2} &= xvi \\ x &= \pm iv \\ \text{If } 10 = 4, y &= two \\ \text{Substitute } (four;2): \quad y &= -2x + c \\ ii &=-2(4) + c \\ 10 &= c \\ y &= -2x + 10 \finish{marshal*}

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Source: https://intl.siyavula.com/read/maths/grade-12/analytical-geometry/07-analytical-geometry-03

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