flflvm's daily question day 2

Another way to do this (more difficult) is to use the  Secant-Tangent Theorem

Note, flflvm97, the line y = x + 3 doesn't actually intersect this circle, but the line y = x + √2 does.....

Call the y coordinate of R, "y".....using symmetry the distance between the two points where the line x = .75 intersects the circle = 2y

And using y = x+ √2, the x coordinate of the intercept of the line and the circle =

x^2 + (x + √2)^2 = 1

x^2 + x^2 + 2√2x + 2 = 1

2x^2 + 2√2x + 1 = 0

2x^2 + √8x + 1 = 0    and using the quad formula x = - √8/4  = -1/√2

And using y = x + √2,   (-1/ √2)^2 + y^2 = 1  so  y^2  = 1/2   and y = the positive root= 1/ √2

So the intersection of the line and the circle is ( -1/√2, 1/√2)

And when x = .75, the y coordinate of this point on the line is  (.75 + √2)

And using the Secant-Tangent Theorem ,we have

( .75 + √2 + y)*(.75 + √2 - y) = (.75 + 1/ √2)^2 + (.75 + √2 - 1/ √2)^2

( .75 + √2 + y)*(.75 + √2 - y) = (.75 + 1/ √2)^2 + (.75 +  1/ √2)^2

(.75 + √2)^2 - y^2  =  2 (.75 + 1/ √2)^2

So

y^2  = (.75 + √2)^2 - 2 (.75 + 1/ √2)^2

And y = ±√((.75 + √2)^2 - 2 (.75 + 1/ √2)^2) = ±.661438 = ±√(7)/4    

And choose the positive root.....and this is the y coordinate of "R".....so R = (3/4,  √(7)/4)

(I told you it was more difficult !!!)

what is this even for math help???

Um...I just want to have fun and motivate.

If you don't like it, pls just pass by.

thank you

The equation of the circle is just   x^2 + y^2 = 1

When x = .75, then y = √(1 - .75^2)  = √(1 - (3/4)^2) = √(1 - 9/16)  = √(7 / 16) = √7 / 4

So, the coordinates of R are (.75, √7 / 4) = (3/4 , √7/4 )

Good one :)

Phil