Question

Draw a circle of radius 4 cm. Construct a pair of tangents to the circle from a point 6 cm away from its centre.

Answer 1

\[ \begin{array}{r l} \textbf{Given:} & \text{A circle of radius } 4 \text{ cm with centre } O, \text{ and a point } P \text{ such that } OP = 6 \text{ cm.} \\[4pt] \textbf{To Find:} & \text{The construction of the pair of tangents from } P, \text{ and the length of each tangent.} \\[4pt] \textbf{Solution:} & \text{Step 1: Draw a circle with centre } O \text{ and radius } 4 \text{ cm.} \\[4pt] & \text{Step 2: Mark a point } P \text{ outside the circle with } OP = 6 \text{ cm and join } OP. \\[4pt] & \text{Step 3: Draw the perpendicular bisector of } OP, \text{ meeting } OP \text{ at its midpoint } M, \text{ so that } MO = MP = 3 \text{ cm.} \\[4pt] & \text{Step 4: With centre } M \text{ and radius } MO = 3 \text{ cm, draw a circle cutting the given circle at } A \text{ and } B. \\[4pt] & \text{Step 5: Join } PA \text{ and } PB; \text{ these are the required tangents.} \\[4pt] & \text{Justification: } OP \text{ is a diameter of the circle with centre } M, \text{ and the angle in a semicircle is a right angle, so } \angle OAP = \angle OBP = 90^\circ. \\[4pt] & \text{Since a line drawn through the end of a radius and perpendicular to it is a tangent, } PA \text{ and } PB \text{ are tangents to the circle.} \\[4pt] & \text{In the right triangle } OAP, \text{ by Pythagoras' theorem:} \\[4pt] & \begin{aligned} PA^{2} &= OP^{2} - OA^{2} \\[4pt] &= 6^{2} - 4^{2} \\[4pt] &= 36 - 16 \\[4pt] &= 20 \\[4pt] PA &= \sqrt{20} \\[4pt] &= 2\sqrt{5} \\[4pt] &= 4.472 \\[4pt] &\approx 4.47 \end{aligned} \\[4pt] \textbf{Answer:} & \text{The length of each tangent is } 2\sqrt{5} \text{ cm} \approx 4.47 \text{ cm.} \end{array} \]