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# Circle tangent

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Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\overline{PA}$ and $\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9.

#2
+19207
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x=AB / 2

tan(APB/2) = 9 / 12 = x / y

y = (12/9) * x

x^2 +y^2 = 12^2

x^2 + [ (12/9) * x ]^2 =12^2

x = 36/5

AB = 2x

AB = 72/5 = 14.4

heureka  Sep 2, 2017
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#1
+7253
0

Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\overline{PA}$ and $\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9.

Right triangle OPA

$$opposite\ side=9\\ adjacent\ side=12\\ tan\ \alpha=\frac{9}{12}\\ \color{blue}\alpha=atan\frac{9}{12}$$

Right triangle PAC

$$hypotenuse=12\\ opposite\ side=\frac{\overline{AB}}{2}\\ \alpha=atan\frac{9}{12}=atan\ 0.75\\ sin\alpha =\frac{\overline{AB}}{2}/12\\\color{blue} \overline{AB}=2\times 12\times sin \alpha=2\times 12\times sin(atan\ 0.75)$$

$$\overline {AB}=14.4$$

!

asinus  Sep 2, 2017
edited by asinus  Sep 2, 2017
edited by asinus  Sep 2, 2017
#2
+19207
+1

x=AB / 2

tan(APB/2) = 9 / 12 = x / y

y = (12/9) * x

x^2 +y^2 = 12^2

x^2 + [ (12/9) * x ]^2 =12^2

x = 36/5

AB = 2x

AB = 72/5 = 14.4

heureka  Sep 2, 2017

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