Higher Maths
The Circle

Let M represent the centre of the circle. M is the midpoint of AB.

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

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

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

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

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:

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

(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.

\(2g=-4\) 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.

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 \(b^2-4ac=0.\)

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

Because the question rules out the situation where the circle passes through the origin and then one point on each of the \(x\)- and \(y\)-axes, we know that the circle intersects one of the axes twice and the other once. This can only occur if the radius is exactly 7 units.

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

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

Note: This answer requires the solution of a quadratic inequality, which is covered within 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 depend on the fact that the radius must be a positive number, so we are going to solve the inequality \(g^2+f^2-c \gt 0\small.\)

But first, we need to define \(g,\) \(f\) and \(c\) in terms of \(p\small.\)

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

Now we can create 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 \(y=(p-1)(5p+2)\small.\)

The 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 \(p\gt 1\small.\)

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 \(b^2-4ac=0.\)

\(2g=-8\) 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:

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

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 \((x-4)^2+(y+2)^2=36.\)

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.

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 \((0,\,-1)\small.\)

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 \((3,\,6)\small.\)

(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:

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

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 \((1,\,2)\small.\)