Tangents drawn from the same external point to a circle have equal lengths. How does this relate $MX$ and $NX$, and how can you use that in the perimeter equation?

Think about the angle between a radius and a tangent at the point of tangency. Which triangle involving $P$, $M$, and $X$ is right, and which side of that triangle is $PX$?

Once you know the lengths of the two legs of the right triangle with hypotenuse $PX$, use the Pythagorean theorem to solve for $PX$.

In Desmos, type the expression (1470 - 2*315)/2 and evaluate it. The result is the common length of the tangent segments $MX$ and $NX$.

Now type sqrt(315^2 + [result]^2) in Desmos, replacing [result] with the value you found for the tangent length (for example, sqrt(315^2 + 420^2)). The output is the length of $PX$; compare this number with the answer choices.

Quadrilateral $PMXN$ has sides $PM$, $MX$, $XN$, and $NP$.

• $PM$ and $PN$ are radii of the circle, so each is $315$.
• $MX$ and $NX$ are tangents drawn from the same external point $X$ to the circle, so $MX = NX$.

Let $MX = NX = t$.

Then the perimeter is

• $PM + MX + XN + NP = 315 + t + t + 315 = 630 + 2t$.

We are told this equals $1{,}470$, so write the equation:

$630 + 2t = 1{,}470$.

From $630 + 2t = 1{,}470$:

1. Subtract $630$ from both sides:

$2t = 1{,}470 - 630 = 840$.

2. Divide both sides by $2$:

$t = 420$.

So $MX = NX = 420$ centimeters.

A radius drawn to a point of tangency is perpendicular to the tangent line.

So in triangle $PMX$:

• $PM$ is a radius, so $PM = 315$.
• $MX$ is a tangent, so $MX = 420$.
• $\triangle PMX$ is a right triangle with a right angle at $M$ (since $PM \perp MX$).

The distance we want, $PX$, is the hypotenuse of this right triangle.

Apply the Pythagorean theorem to right triangle $PMX$:

Compute each square:

• $315^2 = 99{,}225$
• $420^2 = 176{,}400$

Add them:

Now take the square root:

$PX = \sqrt{275{,}625} = 525$.

So the distance between $P$ and $X$ is $525$ centimeters, which corresponds to choice C) 525.