Is Tangent to Circle O at a. If , What Is Myac? (the Figure Is Not Drawn to Scale.)

7.3 Equation of a tangent to a circumvolve (EMCHW)

1. On a suitable system of axes, draw the circle \(x^{2} + y^{ii} = 20\) with center at \(O(0;0)\).
2. Plot the point \(T(2;four)\).
3. Plot the point \(P(0;5)\). Draw \(PT\) and extend the line so that is cuts the positive \(x\)-axis.
4. Mensurate \(O\hat{T}P\).
5. Determine the gradient of the radius \(OT\).
6. Determine the slope of \(PT\).
7. Testify that \(PT \perp OT\).
8. Plot the point \(S(2;-four)\) and join \(Bone\).
9. Draw a tangent to the circumvolve at \(South\).
10. Measure the angle between \(OS\) and the tangent line at \(Southward\).
11. Make a theorize nigh the bending between the radius and the tangent to a circle at a indicate on the circle.
12. Complete the sentence: the product of the \(\ldots \ldots\) of the radius and the slope 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. \(D(ten;y)\) is a signal on the circumference and the equation of the circle is:

\[(10 - a)^{2} + (y - b)^{ii} = r^{2}\]

A tangent is a straight line that touches the circumference of a circumvolve 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 betoken of contact \(D\).

\brainstorm{align*} CD & \perp AB \\ \text{and } C\hat{D}A &= C\hat{D}B = \text{90} ° \finish{marshal*}

The production of the gradient of the radius and the slope 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 circumvolve and write it in the form \[(ten - a)^{2} + (y - b)^{2} = r^{2}\]
2. From the equation, decide the coordinates of the heart of the circle \((a;b)\).
3. Determine the gradient of the radius: \[m_{CD} = \frac{y_{2} - y_{1}}{x_{2}- x_{ane}}\]
4. The radius is perpendicular to the tangent of the circle at a point \(D\) so: \[m_{AB} = - \frac{i}{m_{CD}}\]
5. Write down the gradient-point grade of a direct line equation and substitute \(m_{AB}\) and the coordinates of \(D\). Brand \(y\) the subject field of the equation. \[y - y_{1} = thousand(ten - x_{1})\]

Worked example 12: Equation of a tangent to a circle

Make up one's mind the equation of the tangent to the circle \(10^{2} + y^{2} - 2y + 6x - 7 = 0\) at the point \(F(-two;five)\).

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

Utilise the method of completing the foursquare:

\begin{align*} 10^{2} + y^{2} - 2y + 6x - 7 &= 0 \\ x^{2} + 6x + y^{2} - 2y &= 7 \\ (x^{2} + 6x + 9) - 9 + (y^{2} - 2y + 1) - i &= 7 \\ (x + iii)^{ii} + (y - 1)^{2} &= 17 \end{marshal*}

Draw a sketch

The heart of the circle is \((-iii;1)\) and the radius is \(\sqrt{17}\) units.

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

\begin{align*} m_{CF} &= \frac{y_{2} - y_{1}}{x_{two}- x_{1}}\\ &= \frac{five - 1}{-2 + three}\\ &= 4 \cease{align*}

Determine the gradient of the tangent

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

\begin{align*} m_{CF} \times chiliad &= -1 \\ iv \times m &= -one \\ \therefore one thousand &= - \frac{1}{4} \stop{align*}

Determine the equation of the tangent to the circle

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

\brainstorm{align*} y - y_{1} &= 1000 (x - x_{one}) \\ y - y_{1} &= - \frac{i}{iv} (10 - x_{1}) \\ \text{Substitute } F(-two;5): \quad y - 5 &= - \frac{ane}{4} (x - (-2)) \\ y - five &= - \frac{one}{4} (x + ii) \\ y &= - \frac{1}{4}x - \frac{ane}{2} + 5 \\ &= - \frac{one}{4}x + \frac{9}{2} \end{align*}

Write the terminal answer

The equation of the tangent to the circle at \(F\) is \(y = - \frac{1}{four}10 + \frac{9}{2}\).

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Worked example 13: Equation of a tangent to a circumvolve

The straight line \(y = x + iv\) cuts the circumvolve \(x^{two} + y^{ii} = 26\) at \(P\) and \(Q\).

1. Calculate the coordinates of \(P\) and \(Q\).
2. Sketch the circle and the straight line on the same arrangement 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. Make up one's mind the equations of the tangents to the circle at \(P\) and \(Q\).
6. Decide the coordinates of \(S\), the indicate where the 2 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 \(y = x + iv\) into the equation of the circle and solve for \(ten\):

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

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

Draw a sketch

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

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

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

We need to evidence that the product of the 2 gradients is equal to \(-\text{one}\). From the given equation of \(PQ\), we know that \(m_{PQ} = 1\).

\begin{marshal*} m_{OH} &= \frac{2 - 0}{-2 - 0} \\ &= - 1 \\ & \\ m_{PQ} \times m_{OH} &= - i \\ & \\ \therefore PQ & \perp OH \cease{marshal*}

Make up one's mind the equations of the tangents at \(P\) and \(Q\)

Tangent at \(P\):

Determine the slope of the radius \(OP\):

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

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

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

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

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

Tangent at \(Q\):

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

\begin{align*} m_{OQ} &= \frac{5 - 0}{ane - 0} \\ &= v \cease{align*}

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

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

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

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

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

Decide the coordinates of \(Due south\)

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

\brainstorm{align*} -5x - 26 &= - \frac{one}{5}x + \frac{26}{5} \\ -25x - 130 &= - 10 + 26 \\ -24x &= 156 \\ x &= - \frac{156}{24} \\ &= - \frac{xiii}{2} \\ \text{If } x = - \frac{13}{ii} \quad y &= - v \left( - \frac{13}{2} \right) - 26 \\ &= \frac{65}{two} - 26 \\ &= \frac{xiii}{ii} \end{align*}

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

Show that \(S\), \(H\) and \(O\) are on a straight line

We need to testify that at that place is a constant gradient betwixt whatever 2 of the iii points. We take already shown that \(PQ\) is perpendicular to \(OH\), so we look the gradient of the line through \(Due south\), \(H\) and \(O\) to be \(-\text{one}\).

\begin{marshal*} m_{SH} &= \dfrac{\frac{thirteen}{two} - 2}{- \frac{13}{ii} + 2} \\ &= - 1 \end{align*}\begin{align*} m_{And so} &= \dfrac{\frac{13}{2} - 0}{- \frac{13}{two} - 0} \\ &= - 1 \end{align*}

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

Worked example 14: Equation of a tangent to a circumvolve

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

Depict a sketch

The tangents to the circumvolve, parallel to the line \(y = \frac{ane}{2}x + 1\), must have a gradient of \(\frac{1}{2}\). From the sketch we meet that there are two possible tangents.

Determine the coordinates of \(A\) and \(B\)

To determine the coordinates of \(A\) and \(B\), we must discover the equation of the line perpendicular to \(y = \frac{one}{2}ten + 1\) 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 \(y = - 2x + ane\) into the equation of the circle and solve for \(x\):

\begin{align*} x^{two} + (y-1)^{ii} &= 80 \\ 10^{2} + \left( - 2x + 1 - ane \right)^{two} &= eighty \\ ten^{2} + 4x^{2} &= 80 \\ 5x^{ii} &= lxxx \\ 10^{2} &= sixteen \\ \therefore x &= \pm iv \\ \text{If } x = iv \quad y &= - 2(4) + 1 = - vii \\ \text{If } x = -4 \quad y &= - 2(-4) + i = 9 \end{marshal*}

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

Determine the equations of the tangents to the circle

Tangent at \(A\):

\begin{marshal*} y - y_{1} &= \frac{one}{2} (ten - x_{1}) \\ y - ix &= \frac{1}{2} (x + 4 ) \\ y &= \frac{1}{2} 10 + 11 \end{marshal*}

Tangent at \(B\):

\begin{align*} y - y_{one} &= \frac{1}{2} (x - x_{1}) \\ y + 7 &= \frac{i}{two} (ten - 4 ) \\ y &= \frac{1}{2}x - 9 \finish{marshal*}

The equation of the tangent at point \(A\) is \(y = \frac{ane}{2}x + 11\) and the equation of the tangent at signal \(B\) is \(y = \frac{i}{2}ten - 9\).

Worked example 15: Equation of a tangent to a circle

Determine the equations of the tangents to the circle \(x^{ii} + y^{2} = 25\), from the point \(Thousand(-7;-i)\) outside the circumvolve.

Describe a sketch

Consider where the 2 tangents will touch the circumvolve

Permit the two tangents from \(Yard\) touch the circle at \(F\) and \(H\).

\begin{align*} OF = OH &= \text{five}\text{ units} \quad (\text{equal radii}) \\ OG &= \sqrt{(0 + 7)^{2} + (0 + one)^2} \\ &= \sqrt{l} \\ GF &= \sqrt{ (ten + seven)^{2} + (y + 1)^2} \\ \therefore GF^{2} &= (x + 7)^{two} + (y + i)^2 \\ \text{And } One thousand\hat{F}O = G\hat{H}O &= \text{90} ° \end{align*}

Consider \(\triangle GFO\) and employ the theorem of Pythagoras:

\brainstorm{align*} GF^{two} + OF^{2} &= OG^{ii} \\ \left( x + 7 \right)^{two} + \left( y + one \right)^{2} + v^{ii} &= \left( \sqrt{50} \right)^{2} \\ x^{2} + 14x + 49 + y^{two} + 2y + 1 + 25 &= 50 \\ x^{2} + 14x + y^{2} + 2y + 25 &= 0 \ldots \ldots (one) \\ \text{Substitute } y^{2} = 25 - 10^{2} & \text{ into equation } (1) \\ \quad x^{2} + 14x + \left( 25 - x^{2} \right) + ii\left( \sqrt{25 - 10^{two}} \right) + 25 &= 0 \\ 14x + fifty &= - two\left( \sqrt{25 - x^{two}} \right) \\ 7x + 25 &= - \sqrt{25 - x^{2}} \\ \text{Square both sides: } (7x + 25)^{2} &= \left( - \sqrt{25 - x^{ii}} \right)^{ii} \\ 49x^{2} + 350x + 625 &= 25 - x^{2} \\ 50x^{two} + 350x + 600 &= 0 \\ x^{2} + 7x + 12 &= 0 \\ (x + three)(x + iv) &= 0 \\ \therefore x = -3 & \text{ or } x = -4 \\ \text{At } F: ten = -3 \quad y &= - \sqrt{25 - (-3)^{2}} = - \sqrt{16} = - 4 \\ \text{At } H: x = -four \quad y &= \sqrt{25 - (-4)^{2}} = \sqrt{9} = 3 \end{align*}

Note: from the sketch nosotros see that \(F\) must have a negative \(y\)-coordinate, therefore we have the negative of the foursquare root. Similarly, \(H\) must accept a positive \(y\)-coordinate, therefore nosotros accept the positive of the square root.

This gives the points \(F(-iii;-four)\) and \(H(-4;3)\).

Tangent at \(F\):

\brainstorm{align*} m_{FG} &= \frac{-1 + 4}{-7 + 3} \\ &= - \frac{3}{four} \finish{align*}\brainstorm{marshal*} y - y_{i} &= grand (10 - x_{ane}) \\ y - y_{ane} &= - \frac{3}{4} (x - x_{1}) \\ y + 1 &= - \frac{3}{4} (10 + 7) \\ y &= - \frac{3}{4}x - \frac{21}{4} - i \\ y &= - \frac{iii}{4}x - \frac{25}{four} \terminate{marshal*}

Tangent at \(H\):

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

Write the final reply

The equations of the tangents to the circle are \(y = - \frac{iii}{four}10 - \frac{25}{4}\) and \(y = \frac{4}{iii}x + \frac{25}{3}\).

Equation of a tangent to a circle

Textbook Exercise 7.5

A circumvolve with centre \((eight;-vii)\) and the betoken \((five;-5)\) on the circle are given. Determine the gradient of the radius.

Given:

• the centre of the circle \((a;b) = (viii;-7)\)
• a indicate on the circumference of the circle \((x_1;y_1) = (five;-5)\)

Required:

• the gradient of the radius, \(thousand\)

\begin{align*} one thousand & = \frac{y_2 - y_1}{x_2 - x_1}\\ & = \frac{-5+7}{5-8}\\ & = - \frac{two}{iii} \end{marshal*}

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

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

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

\begin{align*} m_{\bot} & = - \frac{i}{k}\\ & = \frac{-1}{- \frac{two}{three} }\\ & = \frac{iii}{2} \end{align*}

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

Find the gradient of the radius at the bespeak \((2;2)\) on the circumvolve.

Given:

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

Required:

• the slope of the radius, \(g\)

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

Draw a rough sketch:

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

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

Given:

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

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

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

Given a circle with the central coordinates \((a;b) = (-ix;six)\). Determine the equation of the tangent to the circle at the indicate \((-2;5)\).

\begin{marshal*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{v - 6 }{ -2 -(-nine)} \\ & = - \frac{ane}{vii} \stop{align*}

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

\begin{align*} m & = - \frac{i}{m_r} \\ & = \frac{i}{ \frac{one}{vii} } \\ & = 7 \end{align*}

Write downwardly the equation of a directly line and substitute \(1000 = 7\) and \((-2;five)\).

\brainstorm{align*} y_1 & =m x_1 + c\\ 5 & = seven (-2) + c \\ c & = 19 \terminate{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 middle of the circumvolve \(C(a;b) = (1;5)\)
• a bespeak on the circumference of the circumvolve \(H(-2;1)\)

Required:

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

Calculate the gradient of the radius:

\brainstorm{marshal*} m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ & = \frac{i - 5}{-ii - 1 } \\ & = \frac{-4}{-three } \\ & = \frac{4}{3} \finish{align*} \begin{align*} m_r \times chiliad &= -1 \\ m & = - \frac{1}{m_r} \\ & = - \frac{1}{\frac{4}{3} } \\ & = - \frac{3}{4} \terminate{marshal*}

Equation of the tangent:

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

The equation for the tangent to the circle at the point \(H\) is:

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

Given the point \(P(2;-iv)\) on the circle \(\left(ten - four\correct)^{ii} + \left(y + 5\right)^{2} = 5\). Discover the equation of the tangent at \(P\).

Given:

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

Required:

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

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

The gradient of the radius:

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

Equation of the tangent:

\begin{marshal*} y & = m_{\bot} 10 + c\\ -4 & = 2 (ii) + c \\ c & = -8 \finish{align*}

The equation of the tangent to the circle is

\begin{align*} y & = 2 10 - 8 \cease{align*}

Make up one's mind the equation of the circle.

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

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

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

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

The equation of the circle is \(\left(10 + 4\right)^{2} + \left(y - viii\right)^{2} = 136\).

Determine the value of \(m\).

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

\brainstorm{align*} \left(x + 4\correct)^{2} + \left(y - 8\correct)^{2} & = 136 \\ \left(-x + 4\right)^{ii} + \left(m - eight\correct)^{two} & = 136 \\ 36 + \left(k - 8\right)^{2} & = 136 \\ thousand^{ii} - sixteen m + 100 & = 136 \\ m^{two} - sixteen m - 36 & = 0 \\ (m+two)(grand-18) & = 0 \end{align*}

The solution shows that \(y = -2\) or \(y = 18\). From the graph nosotros encounter that the \(y\)-coordinate of \(Q\) must be positive, therefore \(Q(-ten;18)\).

Determine the equation of the tangent to the circle at point \(Q\).

Calculate the gradient of the radius:

\begin{marshal*} m_r & = \frac{y_2 - y_0}{x_2 - x_0} \\ & = \frac{eighteen - 8}{-10 + iv } \\ & = - \frac{10}{6 } \\ & = - \frac{5}{three} \terminate{align*}

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

\brainstorm{align*} m_{\bot} & = - \frac{one}{m_r} \\ & = \frac{one}{\frac{5}{3}} \\ & = \frac{iii}{5} \end{align*}

The equation for the tangent to the circle at the betoken \(Q\) is:

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

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

Substitute the straight line \(y = x + 2\) into the equation of the circumvolve and solve for \(x\):

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

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

Decide the length of \(PQ\).

\begin{align*} PQ &= \sqrt{(x_{ii} - x_{ane})^{two} + (y_{2} - y_{1})^two} \\ &= \sqrt{(-4 -2)^{2} + (-2-iv )^2} \\ &= \sqrt{(-6)^{2} + (-6)^ii} \\ &= \sqrt{36 + 36} \\ &= \sqrt{36 \cdot 2} \\ &= 6\sqrt{2} \end{align*}

Make up one's mind the coordinates of \(Thou\), the mid-indicate of chord \(PQ\).

\begin{marshal*} Yard(10;y) &= \left( \frac{x_{ane} + x_{two}}{2}; \frac{y_{1} + y_{2}}{2} \right) \\ &= \left( \frac{-iv + 2}{2}; \frac{-2 + 4}{2} \correct) \\ &= \left( \frac{-2}{2}; \frac{2}{2} \right) \\ &= \left( -one; 1 \right) \terminate{align*}

If \(O\) is the centre of the circumvolve, show that \(PQ \perp OM\).

\begin{align*} m_{PQ} &= \frac{4 - (-2)}{2 - (-4)} \\ &= \frac{half-dozen}{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_{ii} - y_{1}}{x_{2}- x_{i}} \\ &= \frac{-2 - 0}{- iv - 0} \\ &= \frac{1}{2} \end{align*}

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

\brainstorm{marshal*} m_{OP} \times m_{P} &= -one \\ \frac{1}{2} \times m_{P} &= -i \\ \therefore m_{P} &= - ii \end{align*}

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

\brainstorm{align*} y - y_{1} &= m (ten - x_{i}) \\ y - y_{1} &= - ii (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 slope of the radius \(OQ\):

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

Allow the slope of the tangent at \(Q\) be \(m_{Q}\). The tangent of a circumvolve is perpendicular to the radius, therefore we tin can write:

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

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

\begin{align*} y - y_{1} &= thou (x - x_{1}) \\ y - y_{1} &= - \frac{1}{ii} (10 - x_{1}) \\ \text{Substitute } Q(ii;4): \quad y - 4 &= - \frac{1}{2} (ten - 2) \\ y &= - \frac{1}{2}ten + i + 4 \\ &= - \frac{1}{2}x + five \end{align*}

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

Determine the coordinates of \(S\), the point where the ii tangents intersect.

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

\begin{align*} -2x - 10 &= - \frac{1}{two}x + 5 \\ -4x - xx &= - x + 10 \\ -3x &= xxx \\ x &= - 10 \\ \text{If } x = - x \quad y &= - 2 \left( - 10 \right) - 10 \\ &= 10 \finish{align*}

This gives the indicate \(Due south \left( - 10;ten \right)\).

Show that \(PS = QS\).

\begin{marshal*} PS &= \sqrt{(x_{2} - x_{ane})^{2} + (y_{2} - y_{1})^2} \\ &= \sqrt{(-iv -(-10))^{2} + (-2 - 10)^2} \\ &= \sqrt{(6)^{2} + (-12)^2} \\ &= \sqrt{36 + 144} \\ &= \sqrt{180} \finish{align*} \begin{align*} QS &= \sqrt{(x_{2} - x_{ane})^{two} + (y_{2} - y_{1})^2} \\ &= \sqrt{(2 -(-10))^{2} + (4 - 10)^ii} \\ &= \sqrt{(12)^{2} + (-vi)^2} \\ &= \sqrt{144 + 36} \\ &= \sqrt{180} \terminate{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:

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

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

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