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Question 5
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Consider the following diagram, where BC and BD are tangents to the circle, with centre A. The length of each tangent is 4cm and AB=5cm.

Tangents BC and BD meet the circumference of the circle at C and C respectively. A is the midpoint of the circle with radius AC.



What is the radius of the circle? (Do not write the units.)

Answer 1

To find the radius of the circle, we can use the properties of tangents and right triangles.

Given:
- The lengths of the tangents BC and BD are both 4 cm.
- The distance AB is 5 cm.

Since tangents drawn from a point outside a circle to the circle are equal in length, we know that BC = BD = 4 cm.

Triangle ABC is a right triangle where:
- AC is the radius of the circle (which we need to find).
- AB is the distance from point A (the center) to point B (the point outside the circle) which is 5 cm.
- BC is the tangent from point B to point C, which is 4 cm.

According to the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]

Substituting the known values:
\[ 5^2 = AC^2 + 4^2 \]
\[ 25 = AC^2 + 16 \]

Now, solving for \( AC^2 \):
\[ AC^2 = 25 - 16 \]
\[ AC^2 = 9 \]

Taking the square root:
\[ AC = \sqrt{9} \]
\[ AC = 3 \]

Therefore, the radius of the circle is 3.