We are given a circle with a chord $XY = 16 \, \text{cm}$. The perpendicular bisector of the chord cuts $XY$ at $N$ and the circle at $P$. We also know that $NP = 2 \, \text{cm}$.

Let's break down the problem:

Step 1: Relationship between radius, chord, and bisector

• $N$ is the midpoint of the chord $XY$, so: $XN = \frac{XY}{2} = \frac{16}{2} = 8 \, \text{cm}.$
• The perpendicular bisector passes through the center $O$ of the circle, making $ON$ perpendicular to $XY$.

We are asked to find the radius $r$ of the circle. $O$ is the center of the circle, so we need to calculate $OP$, which is the radius.

Step 2: Right triangle in the circle

• The triangle $ONP$ is a right triangle because $ON$ is perpendicular to $XY$.
• From Pythagoras’ theorem: $OP^2 = ON^2 + NP^2$ where $OP = r$ is the radius we need to find, $ON$ is the distance from the center to the midpoint of the chord, and $NP = 2 \, \text{cm}$ is given.

Step 3: Finding $ON$

To find $ON$, we use the right triangle $ONX$. By Pythagoras’ theorem again: $ON^2 + XN^2 = r^2$ Substitute $XN = 8 \, \text{cm}$: $ON^2 + 8^2 = r^2$ $ON^2 + 64 = r^2$ $ON^2 = r^2 - 64$

Step 4: Applying to $\triangle ONP$

Substituting into the equation for $OP^2$: $OP^2 = ON^2 + NP^2$ $r^2 = (r^2 - 64) + 2^2$ $r^2 = r^2 - 64 + 4$ $r^2 = r^2 - 60$ This simplifies to $r^2 = 60$, so: $r = \sqrt{60} \approx 7.75 \, \text{cm}.$

Final Answer:

The radius of the circle is approximately $7.75 \, \text{cm}$.

Would you like a more detailed breakdown or have any questions? Here are some related questions to explore:

1. How do you find the length of a chord if the radius and the distance from the center to the chord are known?
2. What is the relationship between a chord, its perpendicular bisector, and the radius in a circle?
3. How can you derive the Pythagorean theorem for a right triangle inscribed in a circle?
4. How does the perpendicular bisector theorem help in solving circle-related geometry problems?
5. What are the properties of cyclic quadrilaterals formed by chords and bisectors?

Tip: When dealing with circles, always look for opportunities to apply Pythagoras’ theorem, as many circle problems involve right triangles formed by radii and chords.