Higher Maths
The Circle

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

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

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

Let's just find AM.

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

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

For circle \(C_1\):

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

Let \(r\) represent the radius of \(C_1\).

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

Now, as circle \(C_2\) has the same radius, its equation is \((x-1)^2+(y+4)^2=25.\)

(a)  For circle \(C_1\):

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

For circle \(C_2\):

• 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 \((-g,-f)=(2,-1).\)

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 \(m_{\textsf{tan}}= \large\frac{4}{3}\normalsize.\)

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

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

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:

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

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 \((-g,-f)=(3,7).\)

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 7 units.

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

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

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 \(g^2+f^2-c \gt 0.\)

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

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

Now we can create the inequality:

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

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

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

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 \(p\lt -\small\frac{2}{5}\normalsize\) or \(p\gt 1.\)

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 \(C_1\):

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

Let \(r\) represent the radius of \(C_1\).

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

Now, as circle \(C_2\) has the same radius, its equation is \((x-4)^2+(y+2)^2=36.\)

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

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

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