## Friday, May 21, 2010

### Problem 460: Circle, Tangent, Perpendicular, Radius, Distance, Measurement

Geometry Problem
Click the figure below to see the complete problem 460 about Circle, Tangent, Perpendicular, Radius, Distance, Measurement.

Complete Problem 460

Level: High School, SAT Prep, College geometry

1. Let the circle be y^2 + z^2= x^2, in the y-z plane. Let E be (p, - q). We’ve p^2 + q^2 = x^2, A:(0,x) & B:(x,0). Further, the tangent may be written as: yp-zq-x^2= 0 and a = (- xz –x^2)/(p^2+q^2)^(1/2) = (-xq-x^2)/x =-(q+x). In other words the magnitude of a=(q+x) ignoring its sign. Similarly, b=x – p.
Now RHS = a +b –(2ab)^(1/2)
= (q+x)+(x-p) – (2(q+x)(x-p))^(1/2)
= x+(x+q-p)–(2x^2-2pq-2px+2qx)^(1/2)
= x+(x+q-p)–(x^2+x^2-2pq-2px+2qx)^(1/2)
=x+(x+q-p)–(x^2+p^2+q^2-2pq-2px+2qx)^(1/2)
= x + (x+q-p) – [(x+q-p)^2]^(1/2)
= x + (x+q-p) - (x+q-p)
= x.
Ajit

2. draw OK perpendicular to AF, BL to OE
AK²= x² - ( a - x)² => AK = √ x²-(a-x)
▲AOK ~ ▲BOL ( right tr & OAK = OBL perpendicular sides)
(√2ax -a²)/x = (x-b)/x => x² -(2b + 2a)x + a² + b² = 0
=>
x = a + b - √ 2ab