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

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

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Answer 1 · 2016-12-02T16:32:56

$¯ ¯¯¯¯ ¯ A D = 23.5797$ $bar(AD)=23.5797$

Explanation:

Adopting the origin $(0, 0)$ $(0,0)$ as the common center for $C_{i}$ $C_i$ and $C_{e}$ $C_e$ and calling $r_{i} = 10$ $r_i=10$ and $r_{e} = 16$ $r_e=16$ the tangency point $p_{0} = (x_{0}, y_{0})$ $p_0=(x_0,y_0)$ is at the intersection $C_{i} \cap C_{0}$ $C_i nn C_0$ where

$C_{i} \to x^{2} + y^{2} = r_{i}^{2}$ $C_i->x^2+y^2=r_i^2$
$C_{e} \to x^{2} + y^{2} = r_{e}^{2}$ $C_e->x^2+y^2=r_e^2$
$C_{0} \to {(x - r_{e})}_{e}^{2} + y^{2} = r_{0}^{2}$ $C_0->(x-r_e)^2+y^2= r_0^2$

here $r_{0}^{2} = r_{e}^{2} - r_{i}^{2}$ $r_0^2 = r_e^2-r_i^2$

Solving for $C_{i} \cap C_{0}$ $C_i nn C_0$ we have

${(x^2+y^2=r_i^2),((x-r_e)^2+y^2=r_e^2-r_i^2):}$

Subtracting the first from the second equation

$-2xr_e+r_e^2=r_e^2-r_i^2-r_i^2$ so

$x_0 = r_i^2/r_e$ and $y_0^2= r_i^2-x_0^2$

Finally the sought distance is

$bar(AD)=sqrt((r_e+x_0)^2+y_0^2)=sqrt(r_e^2+3r_i^2)$

or

$bar(AD)=23.5797$

enter image source here

Explanation:

If $bar(BD)$ is tangent to $C_i$ then $hat(ODB) = pi/2$ so we can apply pythagoras:

$bar(OD)^2+bar(DB)^2=bar(OB)^2$ determining $r_0$

$r_0^2= bar(OB)^2-bar(OD)^2=r_e^2-r_i^2$

The point $D$ coordinates, called $(x_0,y_0)$ should be obtained before calculating the sought distance $bar(AD)$

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

$y_0=bar(BD)sin(hat(OBD))$ but $sin(hat(OBD))=bar(OD)/bar(OB)$

then

$y_0 = sqrt(r_e^2-r_i^2)(r_i/r_e)$ and
$x_0=sqrt(r_i^2-y_0^2)$

Answer 2 · 2016-12-10T09:10:19

Drawn

As per given data the above figure is drawn.

O is the common center of two concentric circles

$AB->"diameter of the bigger circle"$

$AO=OB->"radius of the bigger circle"=16 cm$

$DO->"radius of the smaller circle"=10cm$

$BD->"tangent to the smaller circle"->/_BDO=90^@$

Let $/_DOB=theta=>/_AOD=(180-theta)$

In $Delta BDO->cos/_BOD=costheta=(OD)/(OB)=10/16$

Applying cosine law in $Delta ADO$ we get

$AD^2=AO^2+DO^2-2AO*DOcos/_AOD$

$=>AD^2=AO^2+DO^2-2AO*DOcos(180-theta)$

$=>AD^2=AO^2+DO^2+2AO*DOcostheta$

$=>AD^2=AO^2+DO^2+2AO*DOxx(OD)/(OB)$

$=>AD^2=16^2+10^2+2xx16xx10xx10/16$

$=>AD^2=556$

$=>AD=sqrt556=23.58cm$

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

2 Answers

Explanation:

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The radii of two concentric circles are 16 cm and 10 cm. ABABAB is a diameter of the bigger circle. BDBDBD is tangent to the smaller circle touching it at DDD. What is the length of ADADAD?

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