Distance between two intercepts of two tangents to a circle

Question

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.

Answer 1

As mentioned in the body of the question, the equation governing the two tangents to the circle is

$ \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 $

where the equation of the circle is $ (r - H)^T Q (r - H) = 1 $, that is,

$Q = \begin{bmatrix} \dfrac{1}{25} && 0 \\ 0 && \dfrac{1}{25} \end{bmatrix} $

The points on these two tangents that we're interested in lie on the $x$ axis, so their $y$ coordinate is zero, thus the position vector $r = [x, 0]^T $

Further, $A - H = [0, 16]^T - [10, 5]^T = [-10, 11]^T $

and $ r - H = [x, 0]^T - [10, 5]^T = [ x - 10 , -5 ]^T $

Substituting this into the tangents equation, and multiplying through by $(25)^2$, yields,

$ \left( (x - 10)^2 + 25 - 25 \right) \left( 100 + 121 - 25 \right ) = \left( -10(x - 10) - 55 - 25 \right)^2 $

This simplifies to

$ ( x^2 - 20 x + 100 ) ( 196 ) = (-10 x + 20)^2 = 100 x^2 - 400 x + 400$

and this reduces to

$ 96 x^2 - 3520 x + 19200 = 0 $

Dividing through by $32$

$ 3 x^2 - 110 x + 600 = 0 $

The difference in the $x$'s is (from the quadratic equation)

$ d = \Delta x = \dfrac{ \sqrt{ 110^2 - 4(3)(600) } }{ 3 } = \dfrac{ \sqrt{4900}}{3} = \dfrac{70}{3} $

Answer 2

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?

Answer 3

If $\dfrac xa+\dfrac y{16}=1\iff 16x+ay-16a=0$ has to be a tangent of $$(x-10)^2+(y-5)^2=25$$

the distance from the centre = radius

$$5=\dfrac{|16\cdot10+5a-16a|}{\sqrt{16^2+a^2}}$$

$$(160-11a)^2=25(256+a^2)$$

If $a_1,a_2$ are the roots of the quadratic equation mentioned above, we need $|a_2-a_1|=\sqrt{(a_1+a_2)^2-4a_1a_2}$

Can you take it home from here?

Answer 4

Here is an approach using trigonometry. The slope of the angle bisector $AC$ is,

$\tan \alpha = \frac{16 - 5}{0 - 10} = - \frac{11}{10} \implies \sin\alpha = \frac{11}{\sqrt{221}}, \cos\alpha = - \frac{10}{\sqrt{221}}$

$|AC| = \sqrt{(0-10)^2 + (16 - 5)^2} = \sqrt{221}$

$\sin\theta = \frac{5}{\sqrt{221}}, \cos\theta = \frac{14}{\sqrt{221}}$

As $AO = 16$, $$|OP| = - 16 \cot(\alpha - \theta), OR = - 16 \cot(\alpha + \theta)$$

$ \displaystyle d = |PR| = 16 \left[\cot (\alpha - \theta) - \cot (\alpha + \theta) \right]$

$ \displaystyle d = 16 \cdot \frac{2 \sin\theta \cos\theta}{\sin(\alpha + \theta) \sin(\alpha - \theta)}$

$ \displaystyle d = 16 \cdot \frac{2 \cdot 5 \cdot 14 \cdot 221}{(11 \cdot 14 - 10 \cdot 5) (11 \cdot 14 + 10 \cdot 5)} = \frac{70}{3}$

Answer 5

Equation of circle is $x^2 + y^2 – 20x – 10y + 100 = 0$

$AR = $ tangent-length $= \sqrt {[0]^2 + [16]^2 -20[0] – 10[16] + 100} = 14$

$\delta = \tan^{-1} [\dfrac {5}{14}] $

$\beta’ = (\beta’ + \delta) - \delta = tan^{-1} [\dfrac {10}{11}] - \delta$

$\beta = 90^0 - \beta’ =$ a known quantity.

$\alpha = \beta – 2 \times \delta =$ another known quantity.

By surveyor’s formula, $PQ = {16} \times [\dfrac {1}{\tan \alpha} – \dfrac {1}{\tan \beta}]$