Higher Maths
The Circle

\((x-a)^2+(y-b)^2=r^2\) represents a circle centre \((a,\,b)\) and radius \(r\small.\)

So we need to find the centre and radius of this circle.

Let's use C for the centre of the circle. So C is the midpoint of the diameter AB.

$$ \begin{eqnarray} \textsf{C} &=& \left(\small\frac{x_1+x_2}{2}\normalsize,\,\small\frac{y_1+y_2}{2}\normalsize \right) \\[6pt] &=& \left(\small\frac{-3+1}{2}\normalsize,\,\small\frac{8+(-2)}{2}\normalsize \right) \\[6pt] &=& \left(\small\frac{-2}{2}\normalsize,\,\small\frac{6}{2}\normalsize \right) \\[6pt] &=& (-1,\,3) \end{eqnarray} $$

Now we need to find the radius. We could use any of AC, BC or even \(\large\frac{1}{2}\)AB.

$$ \begin{eqnarray} \textsf{AC} &=& \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \\[6pt] &=& \sqrt{(-3-(-1))^2+(8-3)^2} \\[6pt] &=& \sqrt{(-2)^2+5^2}\\[6pt] &=& \sqrt{4+25}\\[6pt] &=& \sqrt{29} \end{eqnarray} $$

So the equation of the circle is \((x+1)^2+(y-3)^2=29\small.\)

Detailed breakdown by ChatGPT 

For circle C₁:

• \(2g=4\) so \(g=2\)
• \(2f=-6\) so \(f=-3\)
• \(c=-12\)
  

Let \(r\) represent the radius of C₁.

$$ \begin{eqnarray} r &=& \sqrt{g^2+f^2-c} \\[6pt] &=& \sqrt{2^2+(-3)^2-(-12)} \\[6pt] &=& \sqrt{4+9+12} \\[6pt] &=& \sqrt{25} \\[6pt] &=& 5 \\[6pt] \end{eqnarray} $$

Now, as circle C₂ has the same radius, its equation is \((x-1)^2+(y+4)^2=25.\)

Side note: We are not fans of the \(g,\,f,\,c\) method, even though the formulae are given. We prefer to convert the expanded form into the standard form by completing the squares. If you want to know how this works, read on:

$$ \begin{gather} x^2+y^2+4x-6y-12=0\\[6pt] (x^2+4x)+(y^2-6y)=12\\[6pt] (x^2\!+\!4x\!+\!4)+(y^2\!-\!6y\!+\!9)=12+4+9\\[6pt] (x+2)^2+(y-3)^2=25\\[6pt] \end{gather} $$

Detailed breakdown by ChatGPT 

(a)  For circle C₁:

• Centre \(= (1,\,-7)\)
• Radius \(= \sqrt{16} = 4\)

For circle C₂:

• Centre \(= (-g,\,-f) = (6,\,5)\)
• Radius \(= \sqrt{g^2+f^2-c}\)
              \(= \sqrt{36+25-12}\)
              \( = \sqrt{49}\)
              \( = 7\)

(b)  We need to find the distance between the centres:

$$ \begin{eqnarray} & & \!\!\!\!\!\!\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\[6pt] &=& \sqrt{(1-6)^2+(-7-5)^2} \\[6pt] &=& \sqrt{(-5)^2+(-12)^2} \\[6pt] &=& \sqrt{25+144}\\[6pt] &=& \sqrt{169}\\[6pt] &=& 13 \end{eqnarray} $$

The sum of the radii is \(4+7=11.\) So as \(13\gt 11\), the circles do not intersect.

Detailed breakdown by ChatGPT 

\(2g=-4\) and \(2f=2\) so the centre C of the circle is \((-g,\,-f)=(2,\,-1)\small.\)

Now we need to find the gradient of the radius CP:

$$ \begin{eqnarray} m_{\textsf{rad}} &=& \small\frac{y_2-y_1}{x_2-x_1}\normalsize \\[6pt] &=& \small\frac{-1-(-4)}{2-6}\normalsize \\[6pt] &=& \small\frac{3}{-4}\normalsize \\[6pt] &=& -\!\small\frac{3}{4}\normalsize \end{eqnarray} $$

The tangent and radius at P are perpendicular, so we use \(m_{\textsf{rad}}\tiny\ \normalsize m_{\textsf{tan}}=-1\) to obtain \(m_{\textsf{tan}}= \large\frac{4}{3}\small.\)

Finally, we use the Nat 5 method to find the equation of the tangent:

$$ \begin{gather} y-b=m(x-a)\\[6pt] y+4=\small{\frac{4}{3}}\normalsize (x-6)\\[6pt] 3(y+4)=4(x-6)\\[6pt] 3y+12=4x-24\\[6pt] 3y=4x-36\\[6pt] \end{gather} $$

Note that alternative forms of the final answer such as \(4x-3y=36\) or \(4x-3y-36=0\) are also acceptable.

Detailed breakdown by ChatGPT 

We substitute the equation of the straight line into the equation of the circle and prove that there is only one solution:

$$ \begin{gather} x^2+y^2-8x+11=0 \\[6pt] x^2+(2x-3)^2-8x+11=0 \\[6pt] x^2+(4x^2-12x+9)-8x+11=0 \\[6pt] 5x^2-20x+20=0 \\[6pt] x^2-4x+4=0 \\[6pt] (x-2)^2=0 \\[6pt] x = 2 \end{gather} $$

There is only one solution, so the line is a tangent.

\(y=2x-3 = 2(2)-3 = 1\) so the point of contact is \((2,1).\)

Note: Instead of factorising the quadratic above and showing that it is a perfect square, we could have shown that the discriminant \(b^2-4ac=0.\)

Detailed breakdown by ChatGPT 

We start by finding the centre of the circle from its equation:

\(2g=-6\) and \(2f=-14\) so the centre is \((-g,\,-f)=(3,\,7)\small.\)

For the circle to be a tangent to one of the axes, the radius must equal 7. This is because the circle with centre \((3,\,7)\) and radius 7 is a tangent to the \(x\)-axis at \((3,\,0)\) and intersects the \(y\)-axis twice, as shown below:

Now that we know the centre and radius, we can obtain the value of \(k\) in either of the following ways.

Method 1: Expand the equation $$ \begin{gather} (x-a)^2+(y-b)^2=r^2\\[6pt] (x-3)^2+(y-7)^2=7^2\\[6pt] (x^2-6x+9)+(y^2-14y+49)=49\\[6pt] x^2+y^2-6x-14y+9=0\\[6pt] \therefore\:k=9\\[6pt] \end{gather} $$

Method 2: Use the radius formula $$ \begin{gather} r=\sqrt{g^2+f^2-k}\\[6pt] 7=\sqrt{(-3)^2+(-7)^2-k}\\[6pt] 7=\sqrt{9+49-k}\\[6pt] 7^2=58-k\\[6pt] 49=58-k\\[6pt] k=9 \end{gather} $$

Detailed breakdown by ChatGPT 

Warning! This solution involves a quadratic inequality, which is part of the Polynomials and Quadratics topic. If you haven't covered that yet, skip this example and come back to it later.

Questions of this type use the fact that \(r=\sqrt{g^2+f^2-c}\,\small.\)

Note that \(g^2+f^2-c\) cannot be negative, as we cannot square root a negative number, or zero, as a circle with zero radius would be just be a dot!

It must therefore be true that \(g^2+f^2-c \gt 0\small.\)

Before solving this inequality, we need to define \(g,\) \(f\) and \(c\) in terms of \(p\small.\) From the given equation:

\(2g=-2p\) so \(g=-p\)
\(2f=-4p\) so \(f=-2p\)
\(c=3p+2\)

Now we can create and solve the inequality:

$$ \begin{gather} g^2+f^2-c\gt 0 \\[6pt] (-p)^2+(-2p)^2-(3p+2)\gt 0 \\[6pt] p^2+4p^2-3p-2\gt 0 \\[6pt] 5p^2-3p-2\gt 0 \\[6pt] (p-1)(5p+2)\gt 0 \\[12pt] \end{gather} $$

The working above gives us "critical values" of \(p=1\) and \(p=-\large\frac{2}{5}\small.\)

In order to state the range of values of \(p\) we can consider the graph of \(y=(p-1)(5p+2)\small.\)

This graph is a "happy" parabola with a minimum turning point between the roots:

So the sections of the graph where \((p-1)(5p+2)\gt 0\) are where \(p\lt -\large\frac{2}{5}\) or \(p\gt 1\small.\)

Detailed breakdown by ChatGPT 

We substitute the equation of the straight line into the equation of the circle and prove that there is only one solution:

$$ \begin{gather} x^2+y^2+2x-4y+5 = 0 \\[6pt] x^2+(3x\!-\!5)^2+2x-4(3x\!-\!5)-5 = 0 \\[6pt] x^2\!+\!9x^2\!-\!30x\!+\!25\!+\!2x\!-\!12x\!+\!20\!-\!5 = 0 \\[6pt] 10x^2-40x+40 = 0 \\[6pt] x^2-4x+4 = 0 \\[6pt] (x-2)^2 = 0 \\[6pt] x = 2 \end{gather} $$

There is only one solution, so the line is a tangent.

\(y=3x-5 = 3(2)-5 = 1\) so the point of contact is \((2,\,1).\)

Note: Instead of factorising the quadratic above and showing that it is a perfect square, we could have shown that the discriminant \(b^2-4ac=0.\)

Detailed breakdown by ChatGPT 

\(2g=-8\) and \(2f=-6\) so the centre C of the circle is \((-g,\,-f)=(4,\,3).\)

Now we need to find the gradient of the radius CP:

$$ \begin{eqnarray} m_{\textsf{rad}} &=& \small\frac{y_2-y_1}{x_2-x_1}\normalsize \\[6pt] &=& \small\frac{3-1}{4-(-2)}\normalsize \\[6pt] &=& \small\frac{2}{6}\normalsize \\[6pt] &=& \small\frac{1}{3}\normalsize \end{eqnarray} $$

The tangent and radius at P are perpendicular, so we use \(m_{\textsf{rad}}\tiny\ \normalsize m_{\textsf{tan}}=-1\) to obtain \(m_{\textsf{tan}}= -3.\)

Finally, we use the Nat 5 method to find the equation of the tangent:

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-1 &=& -\!3(x+2)\\[6pt] y-1 &=& -\!3x-6\\[6pt] y &=& -\!3x-6+1\\[6pt] y &=& -\!3x-5 \end{eqnarray} $$

Detailed breakdown by ChatGPT 

For circle C₁:

• \(2g=-6\) so \(g=-3\)
• \(2f=-2\) so \(f=-1\)
• \(c=-26\)
  

Let \(r\) represent the radius of C₁.

$$ \begin{eqnarray} r &=& \sqrt{g^2+f^2-c} \\[6pt] &=& \sqrt{(-3)^2+(-1)^2-(-26)} \\[6pt] &=& \sqrt{9+1+26} \\[6pt] &=& \sqrt{36} \\[6pt] &=& 6 \\[6pt] \end{eqnarray} $$

Now, as circle C₂ has the same radius, its equation is \((x-4)^2+(y+2)^2=36.\)

Detailed breakdown by ChatGPT 

The length of the square can be determined by subtracting the \(x\)-coordinates of A and D. So the length of the square is \(10-2=8\small.\)

The diameter of each circle is half of the length of the square, so each diameter is \(4\) and each radius is \(2\small.\)

The coordinates of P can easily be found by realising that P is three-quarters of the way along the diagonal AC, ie. \(6\) to the right and \(6\) up from A. So the coordinates of P are \((8,\,7)\small.\)

The equation of the circle with centre P is therefore \((x-8)^2+(y-7)^2=2^2=4\small.\)

Note: There is no need to multiply this out and present the equation in its expanded form.

Detailed breakdown by ChatGPT 

First, we rearrange the equation of the tangent so that we can identify its gradient:

$$ \begin{gather} x+3y=17 \\[6pt] 3y=-x+17 \\[6pt] y=-\small\frac13\normalsize x+\small\frac{17}{3}\normalsize \\[6pt] \end{gather} $$

So \(m_{\textsf{tan}}=-\large\frac13\small,\) and as \(m_{\textsf{rad}}\tiny\ \normalsize m_{\textsf{tan}}=-1\small,\) we know that \(m_{\textsf{rad}}=3\small.\)

We also know that \((2,\,5)\) is on the radius, so its equation is:

$$ \begin{gather} y-5=3(x-2)\\[6pt] y-5=3x-6 \\[6pt] y=3x-1\\[6pt] \end{gather} $$

We are told that the centre is on the \(y\)-axis, ie. the line with equation \(x=0\small,\) so we substitute \(x=0\) to obtain \(y=-1\small.\) The centre of the circle is therefore \((0,\,-1)\small.\)

Detailed breakdown by ChatGPT 

Both equations are true at the point of contact, so we substitute the equation of the straight line into the equation of the circle, solve for \(x\) and then substitute for \(y\small.\)

$$ \begin{gather} x^2+y^2-14x-8y+45=0 \\[6pt] x^2+(2x)^2-14x-8(2x)+45 =0 \\[6pt] x^2+4x^2-14x-16x+45=0 \\[6pt] 5x^2-30x+45=0 \\[6pt] x^2-6x+9=0 \\[6pt] (x-3)^2=0 \\[6pt] x=3 \end{gather} $$

\(y=2x=2\!\times\!3 = 6\) so the point of contact is \((3,\,6)\small.\)

Video solution by Clelland Maths 
Detailed breakdown by ChatGPT 

(a)  For circle C₁:

• \(2g=18\) so \(g=9\)
• \(2f=-2\) so \(f=-1\)
• \(c=-8\)
  

The centre of circle C₁ is therefore \((-g,\,-f) = (-9,\,1)\small.\)

Let \(r\) represent the radius of C₁.

$$ \begin{eqnarray} r &=& \sqrt{g^2+f^2-c}\\[6pt] &=& \sqrt{9^2+(-1)^2-(-8)}\\[6pt] &=& \sqrt{81+1+8}\\[6pt] &=& \sqrt{90}\\[6pt] &=& 3\sqrt{10}\\[6pt] \end{eqnarray} $$

(b)  The two centres are \((-9,\,1)\) and \((-6,\,0)\small.\)

So the distance between the two centres is:

$$ \begin{eqnarray} && \!\!\!\!\!\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\[6pt] &=& \sqrt{(-9-(-6))^2+(1-0)^2}\\[6pt] &=& \sqrt{(-3)^2+1^2}\\[6pt] &=& \sqrt{9+1}\\[6pt] &=& \sqrt{10}\\[12pt] \end{eqnarray} $$

So the radius of C₂ is \(3\sqrt{10}-\sqrt{10}=2\sqrt{10}\small.\)

We now know both the centre and radius of C₂, so we can write its equation:

$$ \begin{gather} (x-(-6))^2+(y-0)^2=\left(2\sqrt{10}\right)^2\\[7pt] (x+6)^2+y^2=40 \end{gather} $$

Video solution by Clelland Maths 
Detailed breakdown by ChatGPT 

To find points of intersection, we substitute the equation of the straight line into the equation of the circle, and solve the resulting quadratic equation:

$$ \begin{gather} x^2+y^2-2x+6y-15=0 \\[6pt] x^2+(x\!+\!1)^2-2x+6(x\!+\!1)-15=0 \\[6pt] x^2\!+\!(x^2\!+\!2x\!+\!1)\!-\!2x+6x\!+\!6-\!15=0 \\[6pt] 2x^2+6x-8=0 \\[6pt] x^2+3x-4=0 \\[6pt] (x+4)(x-1)=0 \\[6pt] x=-4\:\:\small\textsf{or}\normalsize\:\:x=1 \\[6pt] \end{gather} $$

\(x=-4\implies y=\!-4\!+\!1\!=\!-3\small,\) so one point of intersection is \((-4,\,-3)\small.\)

\(x=1\implies y=\!1\!+\!1\!=\!2\small,\) so the other point of intersection is \((1,\,2)\small.\)

Video solution by Clelland Maths 
Detailed breakdown by ChatGPT 

(a)  From \((x+5)^2+(y-6)^2=9\):

Centre of C₁ is \((-5,\,6)\) and radius \(r_1=\sqrt{9}=3.\)

(b)  For C₂:

• \(2g=-14\) so \(g=-7\)
• \(2f=6\) so \(f=3\)
• \(c=54\)

Centre of C₂ is \((-g,\,-f)=(7,\,-3).\)

Let \(r_2\) represent the radius of C₂:

$$ \begin{eqnarray} r_2 &=& \sqrt{g^2+f^2-c} \\[6pt] &=& \sqrt{(-7)^2+3^2-54} \\[6pt] &=& \sqrt{49+9-54} \\[6pt] &=& \sqrt{4} \\[6pt] &=& 2 \end{eqnarray} $$

(c)  This was the final question on Paper 2. It was intended to be unfamiliar and challenging.

First calculate the distance \(d\) between the centres of C₁ \((-5,\,6)\) and C₂ \((7,\,-3)\):

$$ \begin{eqnarray} d &=& \sqrt{(7-(-5))^2+(-3-6)^2} \\[6pt] &=& \sqrt{12^2+(-9)^2} \\[6pt] &=& \sqrt{144+81} \\[6pt] &=& \sqrt{225} \\[6pt] &=& 15 \end{eqnarray} $$

From the diagram, the diameter of C₃ extends from the internal contact point with C₁ to the contact point with C₂.

So the diameter of C₃ is \(r_1 + d - r_2\small,\) which gives us \(3 + 15 - 2 = 16\small.\)

The radius of C₃ is half of this: \(r_3=16\div 2=8\small.\)

The distance from the centre of C₁ \((-5,\,6)\) to the centre of C₃ is \(r_3 - r_1 = 8 - 3 = 5\small.\)

Now we need vector methods to complete the solution.

Since the total distance between the centres of C₁ and C₂ is 15, the centre of C₃ lies \(\frac{5}{15} = \frac{1}{3}\) of the way along the vector from C₁ to C₂. So the position vector of the centre of C₃ is:

$$ \begin{eqnarray} && \begin{pmatrix} -5 \\ 6 \end{pmatrix} + \small\frac{1}{3}\normalsize\begin{pmatrix} 12 \\ -9 \end{pmatrix} \\[8pt] &=& \begin{pmatrix} -5 \\ 6 \end{pmatrix} + \begin{pmatrix} 4 \\ -3 \end{pmatrix} \\[8pt] &=& \begin{pmatrix} -1 \\ 3 \end{pmatrix} \end{eqnarray} $$

So the centre of C₃ is \((-1,\,3)\small,\) and we already know that \(r_3=8\small,\) so we have all of the required information.

Therefore, the equation of circle C₃ is \((x+1)^2+(y-3)^2=64\small.\)

Detailed breakdown by ChatGPT 

\(2g=2\) and \(2f=-4\) so the centre C of the circle is \((-g,\,-f)=(-1,\,2).\)

Now we need to find the gradient of the radius CA:

$$ \begin{eqnarray} m_{\textsf{rad}} &=& \small\frac{y_2-y_1}{x_2-x_1}\normalsize \\[6pt] &=& \small\frac{6-2}{2-(-1)}\normalsize \\[6pt] &=& \small\frac{4}{3}\normalsize \end{eqnarray} $$

The tangent and radius at A are perpendicular, so we use \(m_{\textsf{rad}}\tiny\ \normalsize m_{\textsf{tan}}=-1\) to obtain \(m_{\textsf{tan}}= -\frac{3}{4}.\)

Finally, we use the Nat 5 method to find the equation of the tangent:

$$ \begin{gather} y-b = m(x-a)\\[6pt] y-6 = -\frac{3}{4}(x-2)\\[6pt] 4(y-6) = -3(x-2)\\[6pt] 4y-24 = -3x+6\\[6pt] 3x+4y-30 = 0\\[6pt] \end{gather} $$

Note: You could give your final answer in various equivalent forms, such as \(3x+4y=30\) or \(y = -\frac{3}{4}x+\frac{15}{2}\small.\)

Detailed breakdown by ChatGPT 

Let the centre be \((x_1,\,y_1)=(-2,\,5)\) and the point on the circle be \((x_2,\,y_2)=(3,\,7)\small.\)

We use the distance formula (i.e Pythagoras):

$$ \begin{eqnarray} r^2 &=& (x_2-x_1)^2 + (y_2-y_1)^2 \\[6pt] &=& (3-(-2))^2 + (7-5)^2 \\[6pt] &=& 5^2 + 2^2 \\[6pt] &=& 25 + 4 \\[6pt] &=& 29 \end{eqnarray} $$

Note that there is no need to calculate the radius itself, as the circle equation requires \(r^2\small.\)

Using the centre-radius form \((x-a)^2+(y-b)^2=r^2\) with \((a,\,b)=(-2,\,5)\!\small:\)

$$ (x+2)^2+(y-5)^2=29 $$

Note: We can leave the equation in this form. There is no need to expand the brackets.

Detailed breakdown by ChatGPT