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How would I calculate the radius of the largest circle that will fit within a $90^{\circ}$ sector with radius of $100$ units. (Essentially a quarter of a circle with radius of $100$)?

Do I find the centroid of the sector and calculate the distance from the centroid to the arc? Would the radius of the inner circle be simply half of the the radius of the sector?

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    The center of the circle must lie on the line that bisects the angle. The distance of the center from this angle $=r \sqrt 2, r + r\sqrt 2 \le 100$ solve for $r$ – 2017-02-02
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    I would use an equality $r + r\sqrt 2 =100 \iff r=\dfrac{100}{\sqrt{2}+1}=100(\sqrt{2}-1)\approx 41.4$ – 2017-02-02
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    An interesting variation to the question would be to ascertain the same for a general sector angle $\theta$. – 2017-02-02

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Following @DougM 's comment: Let $C$ the center of the original circle and $D$ the center and $r$ the radius of the circle you are looking for. Now, the wanted circle is tangent to the edges of the orthogonal sector at points let $A, B$. Then you have a square $ACDB$ of edge $r$ with an inscribed orthogonal sector of the circle $(D,r)$. The diagonal $CD$ of the square is $r\sqrt{2}$.

Now extend $CD$ towards $D$ to reach the circle $(D,r)$ at point $E$. You need then that $CD + DE \leq 100$. But $DE=r$ and $CD=r\sqrt{2}$ so you need solve $r(1+\sqrt{2})\leq 100$ and find the maximum value of $r$ that satisfies this.