☰ Contents · Geometry

Practice and the Chapter IV test

Lessons 53–54 · 2 lessons · B. Xaydarov, E. Sariqov, A. Qo‘chqorov. Geometry, Grade 9 (textbook for general secondary schools), revised 4th edition. Huquq va Jamiyat Publishing, Tashkent, 2019
53

Practical exercise and application

Textbook: pp. 140–141
GoalKnow the special cases of the chord and secant theorems: for a point P inside the circle AP·PB = R² − p², outside PA·PB = p² − R²; tangent length √(p² − R²).
New words
p — distance to the centre · p — markazgacha masofaR² − p² · R² − p²p² − R² · p² − R²tangent length · urinma uzunligi
Explanation

Draw the diameter through the point P at distance p from the centre O of a circle of radius R. If P is inside, the diameter splits at P into the parts R − p and R + p, so for any chord AB through P we get AP·PB = (R − p)(R + p) = R² − p². If P is outside, the secant through the centre has PC = p − R and PD = p + R, so for every secant through P we get PA·PB = p² − R². For the tangent, PA² = p² − R², that is PA = √(p² − R²); this also follows from the Pythagorean theorem, as the radius is perpendicular to the tangent. These formulas let us compute an unknown segment directly.

Worked examples
R = 10, p = 6, AP = 4: AP·PB = 100 − 36 = 64, so PB = 16 and AB = 20 (a diameter!).
R = 5, p = 13, PA = 9: PA·PB = 169 − 25 = 144, so PB = 16 and AB = 7.
Class activity

“A rope circle”: mark a circle on the ground with a rope (with an adult, no sharp tools) and measure the products of the parts of two chords through one point inside; check that they are equal.

Practice
1
R = 13, p = 5 (P inside), AP = 9. Find PB.
2
R = 3, p = 5. What is the tangent length?
3
R = 6, p = 10 and on a secant PA = 4. Find AB.
4
Why is the product R² − p² for an interior point?