Maths Coursework - Circle geometry, tangents etc.

Question

This is a screenshot from the harder part of my coursework. Unfortunately I do not understand it and have no one to explain it.

Answer 1

• (a)

The distance $d$, between a point $O(x_0,y_0)$ and a line $ax+by+c=0$ is:

$$d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$

In this case $d=r=4\sqrt5$ and the the line is $x-2y+23=0$, then we get:

$$4\sqrt5={|1x_0-2y_0+23|\over \sqrt{1+4}}$$ $$20=|x_0-2y_0+23|$$ $$x_0-2y_0+23=\pm20$$

and we get two linear equations: $x_0-2y_0=-3$ and $x_0-2y_0=-43$. We choose the first solution because it represents a line (containing the center) under the line $x-2y+23=0$.

• (b)

We can use the distance between two points (Pythagoras's theorem):

$$d^2=r^2=(x_P-x_0)^2+(y_P-y_0)^2$$ $$80=(-7-x_0)^2+(-2-y_0)^2$$

• (c)

\begin{equation} \begin{cases} x_0-2y_0=-3\\(-7-x_0)^2+(-2-y_0)^2=80 \end{cases} \begin{cases} x_0=2y_0-3\\(7+x_0)^2+(2+y_0)^2=80 \end{cases} \begin{cases} x_0=2y_0-3\\(7+2y_0-3)^2+(2+y_0)^2=80 \end{cases} \end{equation}

Now we can solve the second equation $(4+2y_0)^2+(2+y_0)^2=80$:

$$4(2+y_0)^2+(2+y_0)^2=80$$ $$5(2+y_0)^2=80$$ $$2+y_0=\pm4$$

thus we get $y_0=-6$ or $y_0=2$ and $x_0=-15$ or $x_0=1$ respectively. The right center is $O(1,2)$.

• (d)

The equation of a line through $P(-7,-2)$ is: $y+2=m(x+7)$. We can determine $m$ thinking that the line through $P$ must be tangent to the circle, hence normal to the radius; normal lines have to satisfy $m_1m_2=-1$. Let $m_{OP}$ the angular coefficient of the segment $\overline{OP}$, then:

$$m_{OP}=\frac{y_0-y_P}{x_0-x_P}=\frac{2+2}{1+7}={1\over2}$$

thus $m=-2$.

So the line through $P$ is $y+2=-2(x+7)\longrightarrow y=-2x-16$. The point $R$ is obtained by solving the following system:

\begin{equation} \begin{cases} x-2y+23=0\\y=-2x-16 \end{cases} \begin{cases} x+4x+32+23=0\\y=-2x-16 \end{cases} \begin{cases} x=-11\\y=6 \end{cases} \end{equation}

hence $R(-11,6)$.