P1 §6.4 Use tangent and chord properties

Pure mathematics Year 1

Table of contents

1. Introduction
2. Tangent properties
3. Chord properties
4. Edexcel P1 Ch6 Exercise 6E

1. Introduction

We can use the properties of tangents and chords within circles to solve geometric problems.

2. Tangent properties

• A tangent to a circle is a straight line that intersects the circle at only one point.
• A tangent to a circle is perpendicular to the radius of the circle at the point of intersection.

Example 1. The circle $C$ has equation $(x-2)^2+(y-6)^2=100$.
(a) Verify that the point $P(10,0)$ lies on $C$.
(b) Find an equation of the tangent to $C$ at the point $(10,0)$, giving your answer in the form $ax+by+c=0$.

 

Solution.

(a) We verify: $$ \begin{align} (x-2)^2+(y-6)^2&=(10-2)^2+(0-6)^2 \\ &=8^2+(-6)^2 \\ &=64+36 \\ &=100 \quad \checkmark \end{align} $$ (b) We first draw the diagram as shown below.

To find the gradient of the tangent, we need to find the gradient of $PQ$:

$$ \begin{align} m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{0-6}{10-2}=\frac{-6}{8}=-\frac34 \end{align}$$

Since the tangent is perpendicular to $PQ$, its gradient is $\frac43$ and it passes through $P(10,0)$:

$$ \begin{align} y&=\frac43x+c \\ 0&=\frac43\cdot10+c \\ c&=-\frac{40}3 \\ \Rightarrow\qquad y&=\frac43x-\frac{40}3 \\ \Rightarrow\qquad 4x&-3y-40=0 \quad \checkmark \end{align} $$

A similar question may be found in Exercise 6E, Q5.

 

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Example 2. A circle $C$ has equation $(x-5)^2+(y+3)^2=10$.
The line $l$ is a tangent to the circle and has gradient $-3$.
Find two possible equations for $l$, giving your answers in the fom $y=mx+c$.

 

Solution.

Method 1. We can start with sketching a diagram to understand the situation. We then soon realise that there are two possible tangents with the gradient $-3$ as shown below.

Here's our strategy:

• The line $L$ has gradient $\frac13$ since it is perpendicular to $l$, and it passes through the centre $(5,-3)$. So we can find the equation for $L$: $$ \begin{align} y&=\frac13x+c \\ \textrm{Passing through $(5,-3)$}:\qquad -3&=\frac13\cdot5+c \\ \Rightarrow\qquad c&=3+\frac53=-\frac{14}3 \\ \Rightarrow\qquad y&=\frac13x-\frac{14}3 \end{align} $$
• Now, we can find the points of intersection $A$ and $B$ by solving simultaneous equations: $$ \begin{align} (x-5)^2+(y+3)^2&=10 \\ (x-5)^2+\left(\frac13x-\frac{14}3+3\right)^2&=10 \qquad\left(\textrm{since}\quad L:\quad y=\frac13x-\frac{14}3\right) \\ (x-5)^2+\left(\frac13x-\frac53\right)^2&=10 \\ x^2-10x+25+\left(\frac19x^2-\frac{10}9x+\frac{25}9\right)&=10 \\ \frac{10}9x^2-\frac{100}9x+\frac{160}9&=0 \\ x^2-10x+16&=0 \\ (x-2)(x-8)&=0 \\ \Rightarrow\qquad x=2\quad{\rm or}\quad x&=8 \\ \Rightarrow\qquad y=-4\quad{\rm or}\quad y&=-2 \\ \Rightarrow\qquad A(2,-4)\quad{\rm and}\quad B&(8,-2) \end{align} $$
• and hence the two tangents going through $A$ and $B$ are: $$ \begin{align} \textrm{Tangent through } A(2,-4) \; : && y&=-3x+c_1 \\ && -4&=-3\cdot2+c_1 \\ && c_1&=2 \\ && \Rightarrow \qquad y&=-3x+2 \qquad\checkmark \\ \\ \textrm{Tangent through } B(8,-2) \; : && y&=-3x+c_2 \\ && -2&=-3\cdot8+c_2 \\ && c_2&=22 \\ && \Rightarrow\qquad y&=-3x+22 \qquad\checkmark \end{align} $$

 

Method 2. Alternatively, we can use the discriminant to find the values of $c$. It is given that $y=-3x+c$ is a tangent to the circle, i.e. the following simultaneous equations has only one solution.

$$ \left\{\begin{array}{l} (x-5)^2+(y+3)^2=10 \\ y=-3x+c \end{array}\right. $$

As we substitute $y=-3x+c$ into the equation of the circle, we find

$$ \begin{align} (x-5)^2+(y+3)^2&=10 \\ (x-5)^2+(-3x+c+3)^2&=10 \\ x^2-10x+25+9x^2-6(c+3)x+(c+3)^2&=10 \\ 10x^2-2[3(c+3)+5]x+(c+3)^2+15&=0 \end{align} $$

Since it has only one solution, its discriminant must be equal to zero, i.e.

$$ \begin{align} \Delta&=b^2-4ac \\ &=4[3(c+3)+5]^2-4\cdot10\cdot\left[(c+3)^2+15\right] \\ &=4\Big[9(c+3)^2+30(c+3)+25-10(c+3)^2-150\Big] \\ &=4\Big[-(c+3)^2+30(c+3)-125\Big] \\ &=-4\Big[(c+3)^2-30(c+3)+125\Big] \\ &=-4(c+3-25)(c+3-5) \\ &=-4(c-22)(c-2) \\ &=0 \end{align} $$ and we find $c=22$ or $c=2$ as before. Thus, the equations of the two tangents are: $$ \begin{align} \left\{ \begin{array}{l} y=-3x+22 \\ y=-3x+2 \end{array} \right. \qquad\checkmark \end{align} $$

A similar question may be found in Exercise 6E, Q7.

 

Aside. A general case: A circle with equation $$ (x - a)^2 + (x - b)^2 = r^2 $$ The line $l$ is a tangent to the circle and has gradient $m$. Find the two possible equations for $l$ giving in the form $ y = mx + c $.

 

Method 1.

• Let $L$ be the perpendicular line to $l$ through the centre $(a,b)$. It has gradient $-\frac1m$ since it is perpendicular to $l$. Since it passes through the centre $(a,b)$, we can find its equation: $$ \begin{align} y&=-\frac1mx+c \\ \textrm{Passing through $(a,b)$}:\qquad b&=-\frac1m\cdot a+c \\ \Rightarrow\qquad c&=b+\frac{a}{m}\\ \Rightarrow\qquad y&=-\frac1mx+b+\frac{a}{m} \end{align} $$
• Now, we can find the points of intersection A and B by solving simultaneous equations: $$ \begin{align} (x-a)^2+(y-b)^2&=r^2 \\ (x-a)^2+\left(-\frac1mx+b+\frac{a}{m}-b\right)^2&=r^2 \qquad\left(\textrm{since}\quad L:\quad y=-\frac1mx+b+\frac{a}{m}\right) \\ (x-a)^2+\left( - \frac{ x - a }{m} \right)^2&=r^2 \\ (x-a)^2+ \frac{ (x - a)^2 }{m^2 } &=r^2 \\ \left( \frac{1 + m^2}{m^2} \right) (x - a)^2 &= r^2 \\ (x - a)^2 &= \frac{ r^2 m^2 }{ 1 + m^2 } \\ \Rightarrow \qquad x &= a \pm \frac{rm}{ \sqrt{ 1 + m^2 } } \end{align} $$ and the corresponding $y$-coordinates are: $$ \begin{align} y&= - \frac1m \left( a \pm \frac{rm}{ \sqrt{ 1 + m^2 } } \right) + b + \frac{a}{m} \\ &= b \mp \frac{r}{ \sqrt{ 1 + m^2 } } \end{align} $$ So the points of intersection are: $$ \begin{align} A&=\left( a + \frac{rm}{ \sqrt{ 1 + m^2 } }, b - \frac{r}{ \sqrt{ 1 + m^2 } } \right) \\ B&=\left( a - \frac{rm}{ \sqrt{ 1 + m^2 } }, b + \frac{r}{ \sqrt{ 1 + m^2 } } \right) \end{align} $$
• and hence the two tangents going through $A$ and $B$ are: $$ \begin{align} \textrm{Tangent through } A \; : && y&=mx+c_1 \\ && b - \frac{r}{ \sqrt{ 1 + m^2 } } &= m \left( a + \frac{rm}{ \sqrt{ 1 + m^2 } } \right) + c_1 \\ && c_1 &= b - am - r \sqrt{ 1 + m^2 } \\ && \Rightarrow\qquad y&= mx + b - am - r \sqrt{ 1 + m^2 } \qquad\checkmark \\ \\ \textrm{Tangent through } B\; : && y&=mx+c_2 \\ && b + \frac{r}{ \sqrt{ 1 + m^2 } } &= m \left( a - \frac{rm}{ \sqrt{ 1 + m^2 } } \right) + c_2 \\ && c_1 &= b - am + r \sqrt{ 1 + m^2 } \\ && \Rightarrow\qquad y&= mx + b - am + r \sqrt{ 1 + m^2 } \qquad\checkmark \end{align} $$

 

Method 2. It is given that $y = mx + c$ is a tangent to the circle, i.e. the following simultaneous equations has only one solution.

$$ \left\{\begin{array}{l} (x-a)^2+(y-b)^2=r^2 \\ y=mx+c \end{array}\right. $$ As we substitute $y=mx+c$ into the equation of the circle, we find $$ \begin{align} (x-a)^2+(mx+c-b)^2&=r^2 \\ (x-a)^2+(mx+c-b)^2&=r^2 \\ x^2 - 2ax + a^2 + m^2 x^2 + 2m (c-b) x + (c-b)^2 &= r^2 \\ \left( 1 + m^2 \right) x^2 + 2 [ m ( c - b ) - a \big] x + (c-b)^2 + a^2 - r^2 &= 0 \end{align} $$ Since it has only one solution, its discriminant must be equal to zero, i.e. $$ \begin{align} \Delta &= 4 \big[ m (c-b) - a \big]^2 - 4 \left( 1 + m^2 \right) \left[ (c-b)^2 + a^2 - r^2 \right] \\ &= 4 \Big( \big[ m (c-b) - a \big]^2 - \left( 1 + m^2 \right) \left[ (c-b)^2 + a^2 - r^2 \right] \Big) \\ &= 4 \Big[ m^2 (c-b)^2 - 2am (c-b) + a^2 - \left( 1 + m^2 \right) (c-b)^2 - \left( 1 + m^2 \right) a^2 + \left( 1 + m^2 \right) r^2 \Big] \\ &= 4 \Big[ - (c-b)^2 - 2am (c-b) - m^2 a^2 + \left( 1 + m^2 \right) r^2 \Big] \\ &= - 4 \Big[ (c-b)^2 + 2am (c-b) + m^2 a^2 - \left( 1 + m^2 \right) r^2 \Big] \\ &= - 4 \Big[ \big[ (c-b) + am \big]^2 - \left( 1 + m^2 \right) r^2 \Big] \\ &=0 \end{align} $$ and thus we find $$ \begin{align} \Rightarrow \qquad c &= b - am \pm r \sqrt{ 1 + m^2 } \end{align} $$ The equations of the two tangents are: $$ \begin{align} \left\{ \begin{array}{l} y = mx + b - am + r \sqrt{ 1 + m^2 } \\ y = mx + b - am - r \sqrt{ 1 + m^2 } \end{array} \right. \qquad\checkmark \end{align} $$

 

For example,

• $m=0$: the tangents are horizontal lines $$ \begin{align} \left\{ \begin{array}{l} y = b + r \\ y = b - r \end{array} \right. \qquad\checkmark \end{align} $$
• $m=\infty$: the tangents are vertical lines. Rearranging the equation by dividing it by $m$ gives $$ \begin{align} \frac1m y = x + \frac1m b - a \pm \frac{r}{m} \sqrt{ 1 + m^2 } \end{align} $$ As $m \rightarrow \infty$, we find $$ \begin{align} \left\{ \begin{array}{l} x = a + r \\ x = b - r \end{array} \right. \qquad\checkmark \end{align} $$

 

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3. Chord properties

• A chord is a line segment that joins two points on the circumference of a circle.
• The perpendicular bisector of a chord will go through the centre of a circle.

Example 3. The points $P$ and $Q$ lie on a circle with centre $C$, as shown in the diagram.
The point $P$ has coordinates $(-7,-1)$ and the point $Q$ has coordinates $(3,-5)$.
$M$ is the midpoint of the line segment $PQ$.
The line $l$ passes through points $M$ and $C$.
(a) Find an equation for $l$.

Given that the $y$-coordinate of $C$ is $-8$,
(b) show that the $x$-coordinate of $C$ is $-4$
(c)  find an equation of the circle.

 

Solution.

(a) The line $l$ is perpendicular to the segment $PQ$ and passes through $M$.

$$ \begin{align} {\rm Gradient\,of\,PQ}=\frac{\Delta y}{\Delta x}=\frac{-5-(-1)}{3-(-7)}=\frac{-4}{10}=-\frac25 \end{align} $$

and hence the gradient of $l$ is $\frac52$. The coordinate of $M$ is given by

$$ \begin{align} M=\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)=\left(\frac{-7+3}2,\frac{-1-5}2\right)=(-2,-3) \end{align} $$

The equation for $l$ is:

$$ \begin{align} y&=\frac52x+c \\ \textrm{Passing through $M(-2,-3)$:}\qquad -3&=\frac52\cdot(-2)+c \\ c&=2 \\ \Rightarrow\qquad y&=\frac52x+2 \end{align} $$

(b) The centre $C$ is on the line $l$ with the $y$-coordinate $-8$, so its $x$-coordinate can be found by the equation for $l$:

$$ -8=\frac52x+2\qquad\Rightarrow\qquad x=\frac25\cdot(-10)=-4 \quad\checkmark $$

(c) Now, to write down the equation for the circle, we need to find the radius and it can be found by either CP or CQ:

$$ \begin{align} CP&=\sqrt{(-7-(-4))^2+(-1-(-8))^2}=\sqrt{(-3)^2+7^2}=\sqrt{9+49}=\sqrt{58} \\ CQ&=\sqrt{(3-(-4))^2+(-5-(-8))^2}=\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58} \end{align} $$

So the equation of the circle is:

$$ (x+4)^2+(y+8)^2=58 \quad\checkmark $$

A similar question may be found in Exercise 6E, Q10.

 

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4. Edexcel P1 Ch6 Exercise 6E

Question 1. The line $ x + 3y - 11 = 0 $ touches the circle $ (x+1)^2 + (y+6)^2 = r^2$ at $(2,3)$.

(a) Find the radius of the circle.

(b) Show that the radius at $(2,3)$ is perpendicular to the line.

 

Solution.

(a) The point $(2,3)$ is on the circle. $$ \begin{align} r^2 &= (2+1)^2 + (3+6)^2 \\ &= 3^2 + 9^2 \\ &= 90 \\ \Rightarrow \qquad r&= \sqrt{90} = 3\sqrt{10} \end{align} $$

 

Aside. The fact that the line $ x + 3y - 11 = 0 $ touches the circle: $$ \begin{align} (x+1)^2 + (y+6)^2 &= r^2 \\ (-3y+11+1)^2 + (y+6)^2 &= r^2 \\ 9(y-4)^2 + (y+6)^2 &= r^2 \\ 9 \left( y^2 - 8x + 16 \right) + \left( y^2 + 12y + 36 \right) &= r^2 \\ 10 y^2 - 60 y + 180 - r^2 &= 0 \end{align} $$ requires $\Delta = 0$, i.e. $$ \begin{align} \Delta &= b^2 - 4ac \\ &= 60^2 - 4 \cdot 10 \cdot \left( 180 - r^2 \right) \\ &= 40 \left[ 90 - \left( 180 - r^2 \right) \right] \\ &= 40 \left( r^2 - 90 \right) \\ &=0 \\ \Rightarrow \qquad r&= \sqrt{90} = 3\sqrt{10} \qquad \checkmark \end{align} $$ Now, taking $r^2=90$, we can even find the coordinates of the point of intersection: $$ \begin{align} & \Rightarrow & 10 y^2 - 60 y + 90 &= 0 \\ & \Rightarrow & 10 \left( y^2 - 6y + 9 \right) &= 0 \\ & \Rightarrow & 10 \left( y - 3 \right)^2 &= 0 \\ & \Rightarrow & y&= 3 \end{align} $$ and $$ x = -3y + 11 = -3(3)+11 = 2 $$ and thus the point of intersection is at $(2,3)$.

 

(b) The line can be written as $$ y = - \frac13 x + \frac{11}{3} $$ The radius at $(2,3)$ is the line segment joining the point with the centre $(-1,-6)$, i.e. its gradient reads $$ \begin{align} \frac{ \Delta y }{ \Delta x } &= \frac{-6-3}{-1-2} = \frac{-9}{-3}=3 \end{align} $$ Since $m_1m_2=-1$, the radius is perpendicular to the line.

 

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Question 2. The point $(1,-2)$ lies on the circle centre $(4,6)$.

(a) Find the equation of the circle.

(b) Find the equation of the tangent to the circle at $P$.

 

Solution.

(a) The centre of the circle is $(4,6)$: $$ (x-4)^2 + (y-6)^2 = r^2 $$ and the circle passes through the point $(1,-2)$: $$ \begin{align} r^2 &= (1-4)^2 + (-2-6)^2 \\ &= (-3)^2 + (-8)^2 \\ &= 9 + 64 \\ &= 73 \end{align} $$ thus the equation of the circle reads $$ (x-4)^2 + (y-6)^2 = 73 $$

 

(b) The grdient of the radius of the circle at $P(1,-2)$ is: recall the centre $(4,6)$, $$ \begin{align} \frac{ \Delta y }{ \Delta x } = \frac{6-(-2)}{4-1} = \frac{8}{3}. \end{align} $$ Since the tangent at $(1,-2)$ is perpendicular to the radius, the gradient of the tangent is $ - \frac{3}{8}$ and its equation reads $$ y = - \frac{3}{8} x + c $$ Since it passes through $P(1,-2)$, $$ \begin{align} -2 &= - \frac{3}{8} + c \\ \Rightarrow \qquad c &= -2 + \frac{3}{8} = - \frac{13}{8} \end{align} $$ and we find: $$ y = - \frac{3}{8} x - \frac{13}{8} $$

 

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Question 3. The points $A$ and $B$ with coordinates $(-1,-9)$ and $(7,-5)$ lie on the circle $C$ with equation $$ (x - 1)^2 + (y + 3)^2 = 40. $$

(a) Find the equation of the perpendicular bisector of the line segment $AB$.

(b) Show that the perpendicular of bisector $AB$ passes through the centre of the circle $C$.

 

Solution.

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Question 4. (P) The points $P$ and $Q$ with coordinates $(3,1)$ and $(5,-3)$ lie on the circle $C$ with equation $$ x^2 - 4x + y^2 + 4y = 2 $$.

(a) Find the equation of the perpendicular bisector of the line segment $PQ$.

(b) Show that the perpendicular of bisector $PQ$ passes through the centre of the circle $C$.

 

Solution.

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Question 5. (E) The circle $C$ has equation $$ x^2 + 18x + y^2 - 2y + 29 = 0. $$

(a) Verify the point $P(-7,-6)$ lies on $C$. [2 marks]

(b) Find an equation for the tangent to $C$ at the point $P$, giving your answer in the form $ y = mx + c $. [4 marks]

(c) Find the coordinates of $R$, the point of intersection of the tangent and the $y$-axis. [2 marks]

(d) Find the area of the triangle $APR$. [2 marks]

 

Solution.

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Question 6. The tangent to the circle $ (x + 4)^2 + (y - 1)^2 = 242 $ at $(7,-10)$ meets the $y$-axis at $S$ and the $x$-axis at $T$.

(a) Find the coordinates of $S$ and $T$. [5 marks]

(b) Hence, find the area of $ \triangle OST $, where $O$ is the origin. [3 marks]

 

Solution.

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Question 7. The circle $C$ has equation $$ (x + 5)^2 + (y + 3)^2 = 80. $$ The line $l$ is a tangent to the circle and has gradient 2. Find two possible equations for $l$ giving your answers in the form $ y = mx + c $. [8 marks]

 

Solution.

Method 1. Our logic is:

• Since the gradient of the tangent(s) is given, all we need is a point on the tangent(s).
• $P$ and $Q$ are the points on the two tangents.
• They are also the points of intersection between the line $PQ$ and the circle.
• In order to find their coordinates, we need the equation of the line $PQ$.

So, our strategy is the reverse of these, i.e.

• Find the line $PQ$.
• Find the coordinates of $P$ and $Q$, the points of intersection between the line $PQ$ and the circle.
• Using $P$ and $Q$, we can find the equations of the tangents.

(i) The centre of the circle is at $$ A(-5,-3). $$ Since the line $PQ$ and the tangents are perpendicular, the gradient of $PQ$ is $ - \frac12 $. $$ PQ \; : \qquad y = -\frac12 x + e $$ Since it passes through $ A(-5,-3) $, $$ \begin{align} -3 &= \frac{5}{2} + e \\ \Rightarrow \qquad e &= -3 - \frac{5}{2} = - \frac{11}{2} \end{align} $$ and the equation of the line $PQ$ reads $$ y = -\frac12 x - \frac{11}{2} $$

 

(ii) Since $P$ and $Q$ are the points of intersection between the circle and the line $PQ$, the coordinates of $P$ and $Q$ are the solutions of the following simultaneous equations, $$ \begin{align} \left\{ \begin{array}{l} (x + 5)^2 + (y + 3)^2 = 80 \\ y = -\frac12 x - \frac{11}{2} \end{array} \right. \end{align} $$ As we substitute the line equation into the circle equation, we find: $$ \begin{align} && (x + 5)^2 + \left( -\frac12 x - \frac{11}{2} + 3 \right)^2 &= 80 \\ &\Rightarrow& (x + 5)^2 + \left( -\frac12 x - \frac{5}{2} \right)^2 &= 80 \\ &\Rightarrow& (x + 5)^2 + \left( - \frac{ x + 5 }{ 2 } \right)^2 &= 80 \\ &\Rightarrow& (x + 5)^2 + \frac{ ( x + 5 )^2 }{ 4 } &= 80 \\ &\Rightarrow& \frac54 (x + 5)^2 &= 80 \\ &\Rightarrow& (x+5)^2 &= 64 \\ &\Rightarrow& x &= -5 \pm 8 = -13 \quad \textrm{or} \quad 3 \end{align} $$ which are the $x$-coordinate of $P$ and $Q$, respectively. To find $y$-coordinates, use the equation of $PQ$: $ y = -\frac12x - \frac{11}{2}$,i.e.

• For $P$: $x = -13$, $$ \begin{align} y = \frac{13}{2} - \frac{11}{2} = 1 \qquad \Rightarrow \qquad P = (-13,1) \end{align} $$
• For $Q$: $x = 3$, $$ \begin{align} y = -\frac{3}{2} - \frac{11}{2} = - \frac{14}{2} = -7 \qquad \Rightarrow \qquad Q = (3,-7) \end{align} $$

 

(iii) Finally, we can find the equations of the tangents:

• For $l_1$: $y = 2x + c$ passing through $P(-13,1)$. $$ \begin{align} && 1 &= -26 + c \\ &\Rightarrow& c &= 27 \\ \\ &\Rightarrow& y &= 2x + 27 \qquad \checkmark \end{align} $$
• For $l_2$: $y = 2x + d$ passing through $Q(3,-7)$. $$ \begin{align} && -7 &= 6 + d \\ &\Rightarrow& d &= -13 \\ \\ &\Rightarrow& y &= 2x - 13 \qquad \checkmark \end{align} $$

 

Method 2. We can use the discriminant to find the values of $c$, the $y$-intercept of the tangent. It is given that $y=2x+c$ is a tangent to the circle, i.e. the following simultaneous equations has only one solution.

$$ \left\{\begin{array}{l} (x+5)^2+(y+3)^2=80 \\ y=2x+c \end{array}\right. $$

As we substitute $y=2x+c$ into the equation of the circle, we find

$$ \begin{align} && (x+5)^2+(2x+c+3)^2&=80 \\ &\Rightarrow& x^2 + 10x + 25 + 4x^2 + 4(c+3) x + (c+3)^2 &= 80 \\ &\Rightarrow& 5x^2 + 2 \big[ 2(c+3) + 5 \big] x + (c+3)^2 - 55 &= 0 \\ \end{align} $$

Since it has only one solution, its discriminant must be equal to zero, i.e.

$$ \begin{align} \Delta &=4[2(c+3)+5]^2-4 \cdot 5 \cdot \left[(c+3)^2 =- 55 \right] \\ &=4 \Big[ 4 (c+3)^2 + 20(c+3) + 25 - 5 (c+3)^2 + 275 \Big] \\ &=4 \Big[ -(c+3)^2 + 30(c+3) - 300 \Big] \\ &=-4 \Big[(c+3)^2 - 20(c+3) + 300 \Big] \\ &=-4(c+3-30)(c+3+10) \\ &=-4(c-27)(c+13) \\ &=0 \end{align} $$ and we find $c=27$ or $c=-13$. Thus, the equations of the two tangents are: $$ \begin{align} \left\{ \begin{array}{l} y=2x+27 \\ y=2x-13 \end{array} \right. \qquad\checkmark \end{align} $$

 

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Question 8.

 

Solution.

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Question 9. (E/P) The circle $C$ has centre $P(11,-5)$ and passes through the point $Q(5,3)$.
(a) Find an equation for $C$. (3 marks)

The line $l_1$ is a tangent to $C$ at the point $Q$.
(b) Find an equation for $l_1$. (4 marks)

The line $l_2$ is parallel to $l_1$ and passes through the midpoint of $PQ$. Given that $l_2$ intersects $C$ at $A$ and $B$
(c) find the coordinates of points $A$ and $B$ (4 marks)
(d) find the length of the line segment $AB$, leaving your answer in its simplest surd form. (3 marks)

 

Solution.

(a) The centre of the circle is $P(11,-5)$ and the radius is given by $PQ$.

$$ \begin{gather} r=PQ=\sqrt{(5-11)^2+(3-(-5))^2}=\sqrt{(-6)^2+8^2}=10 \\ \therefore\quad (x-11)^2+(y+5)^2=100 \end{gather} $$

(b) The gradient of $PQ$ reads:

$$ \frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-(-5)}{5-11}=\frac{8}{-6}=-\frac43$$

and hence the gradient of $l_1$ is $\frac34$ since it is perpendicular to $PQ$. The equation of the line $l_1$ passing through $Q(5,3)$ is:

$$ \begin{align} y&=\frac34x+c \\ \textrm{Passing through $Q(5,3)$}:\qquad 3&=\frac34\cdot5+c \\ \Rightarrow\qquad c&=-\frac34 \\ \therefore\quad y&=\frac34x-\frac34 \end{align} $$

(c) The midpoint, say $M$, of $PQ$ is given by:

$$ M= \left(\frac{5+11}2,\frac{3-5}2\right)=(8,-1)$$

The equation of the line $l_2$ passing through M is

$$ \begin{align} y&=\frac34x+d \\ \textrm{Passing through $M(8,-1)$}:\qquad -1&=\frac34\cdot8+d \\ \Rightarrow\qquad d&=-7 \\ \Rightarrow\qquad y&=\frac34x-7 \end{align} $$ The points A and B are the points of intersection between the line $l_2$ and the circle $C$, i.e. we solve the simultaneous equations by substituting $l_2$ into $y$ in $C$: $$ \begin{align} (x-11)^2+(y+5)^2&=100 \\ (x-11)^2+\left(\frac34x-7+5\right)^2&=100 \qquad\left(l_2:\quad y=\frac34x-7\right) \\ (x-11)^2+\left(\frac34x-2\right)^2&=100 \\ x^2-22x+121+\left(\frac9{16}x^2-3x+4\right)&=100 \\ \frac{25}{16}x^2-25x+25&=0 \\ x^2-16x+16&=0 \\ (x-8)^2-8^2+16&=0 \\ (x-8)^2&=48 \\ \Rightarrow\qquad x&=8\pm4\sqrt{3} \end{align} $$ The corresponding $y$-values are: $$ \begin{align} y&=\frac34x-7 \\ &=\frac34(8\pm4\sqrt{3})-7 \\ &=6\pm3\sqrt{3}-7 \\ &=-1\pm3\sqrt{3} \end{align} $$ Consider the relative locations of $A$ and $B$, i.e. $x$-coordinate of $B$ is greater than that of $A$, we find $$ \therefore\quad A(8-4\sqrt{3}-1-3\sqrt{3})\qquad{\rm and}\qquad B(8+4\sqrt{3},-1+3\sqrt{3}) $$

(d) The distance $AB$ is given by Pythagoras' theorem:

$$ \begin{align} AB&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ &=\sqrt{\left(8\sqrt{3}\right)^2+\left(6\sqrt{3}\right)^2} \\ &=\sqrt{192+108} \\ &=\sqrt{300} \\ &=10\sqrt{3} \end{align} $$

 

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Question 10.

 

Solution.

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Question 11.

 

Solution.

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Challenge 1.

 

Solution.

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Challenge 2.

 

Solution.