$\overleftrightarrow{wx}$ is tangent to $\odot v$. what is the radius of the circle?

(there is a circle with center v. point w is on the circle, point x is outside the circle. segment wx is 12 cm, segment vx is 37 cm, and segments vw and vx are drawn.)

radius = \boxed{\quad} cm

Question

$\overleftrightarrow{wx}$ is tangent to $\odot v$. what is the radius of the circle?

(there is a circle with center v. point w is on the circle, point x is outside the circle. segment wx is 12 cm, segment vx is 37 cm, and segments vw and vx are drawn.)

radius = \boxed{\quad} cm

Accepted Answer

# Explanation:
## Step1: Apply the tangent - radius theorem
Since $ \overleftrightarrow{WX} $ is tangent to $ \odot V $ at $ W $, then $ VW\perp WX $ (a tangent to a circle is perpendicular to the radius at the point of tangency). So, $ \triangle VWX $ is a right - triangle with $ \angle VWX = 90^{\circ} $, $ VX=37\mathrm{cm} $, $ WX = 12\mathrm{cm} $, and $ VW=r $ (radius of the circle).
## Step2: Use the Pythagorean theorem
In a right - triangle $ a^{2}+b^{2}=c^{2} $, where $ c $ is the hypotenuse and $ a,b $ are the legs. Here, $ c = VX$, $ a=WX$, $ b = VW $. So, $ VW^{2}+WX^{2}=VX^{2}$. Substitute $ VX = 37$, $ WX = 12$ into the formula: $ r^{2}+12^{2}=37^{2}$.
## Step3: Solve for $ r $
$$
\begin{align*}
r^{2}&=37^{2}-12^{2}\
r^{2}&=(37 + 12)(37 - 12)\quad(\text{using }a^{2}-b^{2}=(a + b)(a - b))\
r^{2}&=(49)(25)\
r^{2}&=1225\
r&=\sqrt{1225}\
r& = 35
\end{align*}
$$

# Answer:
$35$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the tangent - radius theorem

Step2: Use the Pythagorean theorem

Step3: Solve for \( r \)

Answer: