Let's break down this geometry problem step by step. We are given a circle with a specific radius and a point outside the circle from which a tangent is drawn. We need to find the length of this tangent segment.

Here's what we know:

• The radius of the circle (\(r\)) is 5 cm.
• The distance of the external point from the centre of the circle (\(d\)) is 10 cm.

We need to find the length of the tangent drawn from this point to the circle.

Understanding the Geometry of Tangents

A crucial property related to circles and tangents is that a tangent at any point of a circle is perpendicular to the radius through the point of contact. This is a fundamental theorem in circle geometry.

Consider the following:

• Let O be the centre of the circle.
• Let P be the external point from which the tangent is drawn.
• Let T be the point where the tangent touches the circle (the point of contact).

The line segment OT is the radius, so \(OT = r = 5\) cm. The distance from the centre O to the external point P is given as \(OP = d = 10\) cm. The line segment PT is the tangent segment whose length we need to find.

According to the property mentioned, the radius OT is perpendicular to the tangent PT at the point of contact T. This means that the angle \(\angle OTP\) is a right angle (\(90^\circ\)).

Thus, the points O, T, and P form a right-angled triangle, \(\triangle OTP\), with the right angle at T.

Applying the Pythagorean Theorem to Find Tangent Length

In the right-angled triangle \(\triangle OTP\), the side opposite the right angle is the hypotenuse. In this case, the hypotenuse is OP, which is the distance from the center to the external point. The other two sides are the radius OT and the tangent length PT.

According to 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.

For \(\triangle OTP\), the theorem states:

\[ OP^2 = OT^2 + PT^2 \]

We are given \(OP = 10\) cm and \(OT = 5\) cm. Let the length of the tangent \(PT\) be \(t\). Substituting the values into the equation:

\[ 10^2 = 5^2 + t^2 \]

Now, we can solve for \(t\):

\[ 100 = 25 + t^2 \]

Subtract 25 from both sides:

\[ t^2 = 100 - 25 \]

\[ t^2 = 75 \]

To find \(t\), we take the square root of 75:

\[ t = \sqrt{75} \]

We can simplify the square root of 75 by factoring 75 into its prime factors or finding a perfect square factor:

\[ 75 = 25 \times 3 \]

So,

\[ t = \sqrt{25 \times 3} \]

Using the property \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\):

\[ t = \sqrt{25} \times \sqrt{3} \]

Since \(\sqrt{25} = 5\):

\[ t = 5 \times \sqrt{3} \]

\[ t = 5\sqrt{3} \text{ cm} \]

So, the length of the tangent drawn to the circle from the given point is \(5\sqrt{3}\) cm.

Summary of Calculation

Revision Table: Circle Tangent Properties & Formulas

Additional Information on Circles and Tangents

Understanding tangents is key in circle geometry. Here are some related points:

• A tangent intersects the circle at exactly one point (the point of contact).
• From an external point, exactly two tangents can be drawn to a circle.
• The lengths of the two tangents drawn from an external point to a circle are equal. In our setup, if we drew another tangent from P to the circle touching at T', then \(PT = PT'\).
• A line that intersects a circle at two distinct points is called a secant.
• The segment of a secant inside the circle is a chord.

These concepts often appear together in problems involving circles.