Length of Shadow from a lamp?

Question

In the coordinate plane, there is a lamp at point $(0,8).$ There is also a solid circle of radius 1, centered at $(4,5).$ The solid circle casts a shadow on the $x$-axis, as shown. How long is the shadow?

Answer 1

Let $y=mx+8$ be a tangent to the circle. The distance from $(4,5)$ to this line is

$$\frac{|m(4)-(5)+8|}{\sqrt{m^2+1}}=\frac{|m+3|}{\sqrt{m^2+1}}$$

So, we have

\begin{align*} \frac{|m+3|}{\sqrt{m^2+1}}&=1\\ (4m+3)^2&=m^2+1\\ 15m^2+24m+8&=0 \end{align*}

The two values of $m$ are the slopes of the two tangents. Let them be $m_1$ and $m_2$.

The $x$-intercepts are then $\frac{-8}{m_1}$ and $\frac{-8}{m_2}$.

\begin{align*} \left(\frac{-8}{m_1}-\frac{-8}{m_2}\right)^2&=\frac{64[(m_1+m_2)^2-4m_1m_2]}{(m_1m_2)^2}\\ &=\frac{64[(\frac{-24}{15})^2-4(\frac{8}{15})]}{(\frac{8}{15})^2}\\ &=96 \end{align*}

The length of the shadow is $4\sqrt{6}$.

Answer 2

Hint: The distance between a line $ax + by + c = 0$ and a point $(x_0, y_0)$ is $\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$. You also know that the distance between the centre of a circle and its tangent lines is the radius. You can use this to find the tangent lines.

Answer 3

HINT

1. Write down the equaton for the lines passin through $(0,8)$: $y-8=m (x-0)\implies y=mx+8$.
2. Write down the equaton for the circle centered at $(4,5)$ with radius 1: $(x-4)^2+(y-5)^2=1$.
3. Find the two tangent lines (plug $y=mx+8$ in the circle equation and set $\Delta=0$ to find the two values for m).
4. Find the coordinates of the two lines for y=0 and then the length.