From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
Step-by-Step Solution
Key Concept: The tangent drawn from an external point to a circle is perpendicular to the radius at the point of contact. Hence, the line joining the centre O, the external point Q and the point of tangency T forms a right‑angled triangle with OT as one side (radius), QT as the tangent, and OQ as the hypotenuse. Use the Pythagorean theorem.
1. Draw the figure: Let O be the centre of the circle, Q the external point and T the point of tangency. Then \(OT \perp QT\) and \(OQ = 25\) cm, \(QT = 24\) cm.
2. Identify the right‑angled triangle: Triangle \(\triangle OQT\) is right‑angled at T.
3. Apply Pythagoras theorem:
$$OQ^{2}=OT^{2}+QT^{2}$$
Substituting the known values:
$$25^{2}=r^{2}+24^{2}$$
$$625 = r^{2}+576$$
4. Solve for the radius \(r\):
$$r^{2}=625-576=49$$
$$r=\sqrt{49}=7\text{ cm}$$
5. Select the correct option: The radius is 7 cm, which corresponds to option (A).
Correct Answer: 7 cm (Option A)