Question

ABCDE is a circle centre O. The diameter, AC, is extended to the point F so that CF = 16 cm. The line BF is the tangent to the circle at B and FDE is a straight line such that FD = 18 cm and DE = 14 cm. The radius of the circle is r

Answer 1

(a) FB = 24 cm

(b) r = 10

Explanation:

Part (a)

Find the length, in cm, of FB.

Intersecting Secant and Tangent Theorem

When a secant segment and a tangent segment meet at an exterior point, the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

Given:

• Tangent segment = FB
• Secant segment = FE = 18 + 14 = 32 cm
• External secant segment = FD = 18 cm

`\begin{aligned}\implies FB^2 & =FE \cdot FD\\FB^2&=32 \cdot 18\\FB^2&=576\\√(FB^2)&=√(576)\\FB&=24\end{aligned}`

Therefore, the length of FB is 24 cm.

Part (b)

Find the value of r.

Triangle FBO is a right triangle with side lengths:

• BO = r cm
• FB = 24 cm
• OF = (r + 16) cm

Pythagoras Theorem

`a^2+b^2=c^2`

where:

• a and b are the legs of the right triangle.
• c is the hypotenuse (longest side) of the right triangle.

Therefore, substitute the side lengths of triangle FBO into Pythagoras Theorem and solve for r:

`\begin{aligned}\implies BO^2+FB^2&=OF^2\\r^2+24^2&=(r+16)^2\\r^2+576&=r^2+32r+256\\576&=32r+256\\32r&=320\\r&=10\end{aligned}`

Therefore, the value of r is 10.

Answer 2

• r = 10 cm

Explanation:

It is assumed you are looking for the missing value of r.

Use the intersecting secants theorem:

• 
  `FA*FC = FE*FD`

Substitute known values and solve for r:

• (2r + 16)*16 = (14 + 18)*18
• 32r + 256 = 576
• 32r = 320
• r = 10