Out of two concentric circles, the radius of the outer circle is 5 cm and the chord \(AC\) of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
Inner radius = \(3\,\text{cm}\).
Step 1: We have two circles with the same center (concentric). Outer circle radius = \(R = 5\,\text{cm}\). A chord \(AC\) of the outer circle has length 8 cm, and this chord just touches (is tangent to) the inner circle.
Step 2: If a chord of the outer circle is at a distance \(d\) from the center, then the formula for the chord length is: \[ L = 2\sqrt{R^2 - d^2} \] where: • \(L\) = chord length, • \(R\) = radius of outer circle, • \(d\) = perpendicular distance from center to chord.
Step 3: In our case, the distance \(d\) is also the radius of the inner circle (because the chord is tangent to the inner circle). So, let inner radius = \(r\). That means \(d = r\).
Step 4: Substitute the known values into the chord formula: \[ 8 = 2\sqrt{5^2 - r^2} \]
Step 5: Simplify step by step: \[ 8 = 2\sqrt{25 - r^2} \] Divide both sides by 2: \[ 4 = \sqrt{25 - r^2} \]
Step 6: Square both sides to remove the square root: \[ 4^2 = 25 - r^2 \] \[ 16 = 25 - r^2 \]
Step 7: Rearrange to find \(r^2\): \[ r^2 = 25 - 16 = 9 \]
Step 8: Take square root: \[ r = \sqrt{9} = 3 \]
Final Answer: The radius of the inner circle is \(3\,\text{cm}\).