Q6:

The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.

Given:

1. A circle with center $O$.

2. A point $A$ outside the circle such that the distance from the center $O$ to point $A$ is $OA = 5$ cm.

3. A tangent $AB$ from point $A$ to the circle at point $B$, where the length of the tangent $AB = 4$ cm.

To find:

The radius of the circle, denoted as $OB$.

Step 1: Applying the Tangent-Radius Theorem

According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OB \perp AB$. This implies that $\angle OBA = 90^\circ$.

Step 2: Identifying the Geometric Figure

Since $\angle OBA = 90^\circ$, the triangle $\triangle OBA$ is a right-angled triangle, where $OA$ is the hypotenuse, $OB$ is the radius (base/height), and $AB$ is the tangent.

Step 3: Applying the Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [Using Pythagoras Theorem: $Hypotenuse^2 = Base^2 + Height^2$]

$OA^2 = OB^2 + AB^2$

Step 4: Substituting the Given Values

Given $OA = 5$ cm and $AB = 4$ cm.

$(5)^2 = OB^2 + (4)^2$

Step 5: Solving for the Radius ($OB$)

$25 = OB^2 + 16$

Subtract 16 from both sides:

$OB^2 = 25 - 16$

$OB^2 = 9$

Taking the square root on both sides:

$OB = \sqrt{9}$

$OB = 3$ cm

[Since length cannot be negative, we take the positive root.]

Final Answer: The radius of the circle is 3 cm.