The radii of two concentric circles are 16 cm and 10 cm. AB is a diameter of the bigger circle. BD is tangent to the smaller circle touching it at D. What is the length of AD?

Dec 2, 2016

$\overline{A D} = 23.5797$

Explanation:

Adopting the origin $\left(0 , 0\right)$ as the common center for ${C}_{i}$ and ${C}_{e}$ and calling ${r}_{i} = 10$ and ${r}_{e} = 16$ the tangency point ${p}_{0} = \left({x}_{0} , {y}_{0}\right)$ is at the intersection ${C}_{i} \cap {C}_{0}$ where

${C}_{i} \to {x}^{2} + {y}^{2} = {r}_{i}^{2}$
${C}_{e} \to {x}^{2} + {y}^{2} = {r}_{e}^{2}$
${C}_{0} \to {\left(x - {r}_{e}\right)}^{2} + {y}^{2} = {r}_{0}^{2}$

here ${r}_{0}^{2} = {r}_{e}^{2} - {r}_{i}^{2}$

Solving for ${C}_{i} \cap {C}_{0}$ we have

$\left\{\begin{matrix}{x}^{2} + {y}^{2} = {r}_{i}^{2} \\ {\left(x - {r}_{e}\right)}^{2} + {y}^{2} = {r}_{e}^{2} - {r}_{i}^{2}\end{matrix}\right.$

Subtracting the first from the second equation

$- 2 x {r}_{e} + {r}_{e}^{2} = {r}_{e}^{2} - {r}_{i}^{2} - {r}_{i}^{2}$ so

${x}_{0} = {r}_{i}^{2} / {r}_{e}$ and ${y}_{0}^{2} = {r}_{i}^{2} - {x}_{0}^{2}$

Finally the sought distance is

$\overline{A D} = \sqrt{{\left({r}_{e} + {x}_{0}\right)}^{2} + {y}_{0}^{2}} = \sqrt{{r}_{e}^{2} + 3 {r}_{i}^{2}}$

or

$\overline{A D} = 23.5797$

Explanation:

If $\overline{B D}$ is tangent to ${C}_{i}$ then $\hat{O D B} = \frac{\pi}{2}$ so we can apply pythagoras:

${\overline{O D}}^{2} + {\overline{D B}}^{2} = {\overline{O B}}^{2}$ determining ${r}_{0}$

${r}_{0}^{2} = {\overline{O B}}^{2} - {\overline{O D}}^{2} = {r}_{e}^{2} - {r}_{i}^{2}$

The point $D$ coordinates, called $\left({x}_{0} , {y}_{0}\right)$ should be obtained before calculating the sought distance $\overline{A D}$

There are many ways to do that. An alternative method is

${y}_{0} = \overline{B D} \sin \left(\hat{O B D}\right)$ but $\sin \left(\hat{O B D}\right) = \frac{\overline{O D}}{\overline{O B}}$

then

${y}_{0} = \sqrt{{r}_{e}^{2} - {r}_{i}^{2}} \left({r}_{i} / {r}_{e}\right)$ and
${x}_{0} = \sqrt{{r}_{i}^{2} - {y}_{0}^{2}}$

Dec 10, 2016

As per given data the above figure is drawn.

O is the common center of two concentric circles

$A B \to \text{diameter of the bigger circle}$

$A O = O B \to \text{radius of the bigger circle} = 16 c m$

$D O \to \text{radius of the smaller circle} = 10 c m$

$B D \to \text{tangent to the smaller circle} \to \angle B D O = {90}^{\circ}$

Let $\angle D O B = \theta \implies \angle A O D = \left(180 - \theta\right)$

In $\Delta B D O \to \cos \angle B O D = \cos \theta = \frac{O D}{O B} = \frac{10}{16}$

Applying cosine law in $\Delta A D O$ we get

$A {D}^{2} = A {O}^{2} + D {O}^{2} - 2 A O \cdot D O \cos \angle A O D$

$\implies A {D}^{2} = A {O}^{2} + D {O}^{2} - 2 A O \cdot D O \cos \left(180 - \theta\right)$

$\implies A {D}^{2} = A {O}^{2} + D {O}^{2} + 2 A O \cdot D O \cos \theta$

$\implies A {D}^{2} = A {O}^{2} + D {O}^{2} + 2 A O \cdot D O \times \frac{O D}{O B}$

$\implies A {D}^{2} = {16}^{2} + {10}^{2} + 2 \times 16 \times 10 \times \frac{10}{16}$

$\implies A {D}^{2} = 556$

$\implies A D = \sqrt{556} = 23.58 c m$