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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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একটি বৃত্তে, ব্যাস PQ কে Q বিন্দুর মধ্য দিয়ে একটি বহিঃস্থ বিন্দু X পর্যন্ত প্রসারিত করা হয়। বৃত্তের Y বিন্দুতে স্পর্শক XY অঙ্কন করা হয়। যদি XY = 15 সেমি এবং QX = 9 সেমি হয়, তাহলে বৃত্তের ব্যাসার্ধ নির্ণয় করুন।
7 সেমি
10 সেমি
8 সেমি
9 সেমি
8 সেমি
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.