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Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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ஒரு வட்டத்தில், விட்டம் PQ ஆனது Q வழியாக வெளிப்புள்ளி X வரை நீட்டிக்கப்படுகிறது. XY தொடுகோடு வட்டத்தின் புள்ளி Y இல் வரையப்பட்டுள்ளது. XY = 15 செ.மீ மற்றும் QX = 9 செ.மீ எனில், வட்டத்தின் ஆரம் காண்க.
7 செ.மீ
10 செ.மீ
8 செ.மீ
9 செ.மீ
8 செ.மீ
Use the Tangent-Secant theorem: the square of the tangent equals the product of the external segment and the full secant length. Here, 152=9×(9+2r), so 225=81+18r, leading to 144=18r.
Diameter PQ is extended to point X. QX = 9 cm. Tangent XY = 15 cm. We need the radius r.
XY2=QX×PX
Use the Tangent-Secant theorem: the square of the tangent equals the product of the external segment and the full secant length. Here, 152=9×(9+2r), so 225=81+18r, leading to 144=18r.
Many students mistake the secant segment PX as just XQ, forgetting that the secant length must include the diameter PQ (2r).
Identify Tangent-Secant Theorem
For a tangent segment XY and a secant segment XP passing through the center of the circle, the theorem states that the square of the tangent segment equals the product of the external part and the whole secant.
XY2=QX×PX
Substitute Known Values
We know XY=15, QX=9. The full secant length PX=QX+PQ=9+2r, where r is the radius.
152=9×(9+2r)
Solve for Radius r
Expanding the equation: 225=81+18r. Subtracting 81 from both sides gives 144=18r. Dividing by 18 gives r=8.
r=18144=8 cm
C is correct because calculating the radius using the Tangent-Secant theorem 152=9(9+2r) yields r = 8 cm.
This theorem is a specific case of the Power of a Point theorem. Remember that for any point outside a circle, the power is constant for all secants/tangents, a concept useful in Coordinate Geometry.