In right triangle $\triangle OMA$, you know both legs: $OM$ (given) and $AM$ (half the chord). Use the Pythagorean theorem.

The angle formed by two tangents from an external point is $180^\circ$ minus the central angle subtending the same chord.

Once you know $\angle APB$ and that $PA=PB$, decide what kind of triangle $\triangle APB$ is and relate its legs to its hypotenuse $AB$. Рrо⁠pertу‌ οf ⁠Aniкo.ai

Place the center at the origin: $O=(0,0)$. Make the chord horizontal at distance 5 from the origin by using the line $y=5$.

Since $r^2=5^2+5^2=50$, graph the circle $x^2+y^2=50$ and the line $y=5$. Click the intersection points; they should be $A=(5,5)$ and $B=(-5,5)$.

A tangent to $x^2+y^2=50$ at $(x_0,y_0)$ has equation $x x_0 + y y_0 = 50$.

Enter the two tangent lines:

• At $(5,5)$: $5x+5y=50$ (equivalently $x+y=10$)

• At $(-5,5)$: $-5x+5y=50$ (equivalently $-x+y=10$)

Click their intersection to get point $P$. Cο‍nte⁠nt bу Аniкo.а​i

Use the distance formula with the coordinates shown by Desmos:

The value you get is the correct choice.

Let $M$ be the midpoint of chord $\overline{AB}$. Then $AM=\frac{10}{2}=5$ and $OM=5$.

Since $OM\perp AB$, triangle $\triangle OMA$ is right, so

Thus $OA=r=5\sqrt{2}$.

In right triangle $\triangle OMA$, the legs are equal ($OM=AM=5$), so it is a $45$-$45$-$90$ triangle. Therefore,

Because $M$ lies on chord $AB$, ray $OM$ bisects the central angle, so (Аniko⁠.аі)

For two tangents from an external point $P$, the angle between them satisfies

So $\angle APB = 180^\circ - 90^\circ = 90^\circ$.

Also, tangents from the same external point have equal lengths, so $PA=PB$. Thus $\triangle APB$ is a right isosceles triangle with hypotenuse $AB=10$.

In a $45$-$45$-$90$ triangle, each leg equals $\frac{\text{hypotenuse}}{\sqrt{2}}$. Therefore,

So the correct choice is $5\sqrt{2}$.