+0

# Circle tangent

+1
67
2
+148

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
+18566
+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
Sort:

#1
+6931
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
+18566
+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

### 28 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details