Geometry Problem

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

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Complete Problem 460

Level: High School, SAT Prep, College geometry

## Friday, May 21, 2010

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

Labels:
circle,
distance,
measurement,
perpendicular,
radius,
tangent

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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.

ReplyDeleteNow 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

draw OK perpendicular to AF, BL to OE

ReplyDeleteAK²= 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

Please add in my first comment

ReplyDeleteIf E would be between A and B than

x = a + b + √2ab ( from the same formula )