A circle of radius $5$ is centered at $H(10,5)$. Tangents from $A(0, 16) $ are drawn to the circle as shown in the diagram below. Find the distance $d$ between their $x$-axis intercepts.
Here is what I have tried:
If we let $ r = [x, y]^T $, then the equation of both tangents is given by
$ \left( (r – H)^T Q (r – H) – 1 \right) \left( (A – H)^T Q (A – H) – 1 \right) = \left( (r – H)^T Q (A – H) – 1 \right)^2 $
This equation is similar to one found here
where $Q = \begin{bmatrix} \dfrac{1}{25} && 0 \\ 0 && \dfrac{1}{25} \end{bmatrix} $
From here, and since we're interested in the $x$-intercepts, then we want to set $y = 0$, hence $r = [x, 0]^T $
Substituting this into the above equation yields a quadratic equation in $x$ from which the difference in the two solutions gives the required distance.
Best Answer
HINT
We are interested in the lines that pass through $A = (0,16)$ and are tangent to the proposed circle.
Let us denote them generically by $y = mx + 16$. The equation of the circle is given by: \begin{align*} (x - 10)^{2} + (y - 5)^{2} = 25 \end{align*}
Consequently, we are interested in the values of $m$ so that the equation \begin{align*} (x - 10)^{2} + (mx + 11)^{2} = 25 & \Longleftrightarrow (m^{2} + 1)x^{2} + (22m - 20)x + 196 = 0 \end{align*}
has an unique root. This means that $\Delta = 0$. More precisely, \begin{align*} (22m - 20)^{2} -784(m^{2} + 1) = -300m^{2} - 880m - 384 = 0 \end{align*}
Once you have the values of $m$, you also have the expressions of the desired lines.
Can you take it from here?