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# flflvm's daily question day 2

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flflvm's daily question is here!!

Here you go, hope you guys enjoy it.

Line PQ is tangent to the circle. Mid-point of circle is 0,0. Equation of the Line PQ is y=x+3. Find the co-ordinates of point R on equation x=0.75.

flflvm97  Dec 12, 2014

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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 !!!)

CPhill  Dec 13, 2014
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what is this even for math help???

jlara1017  Dec 12, 2014
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Um...I just want to have fun and motivate.

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

thank you

flflvm97  Dec 12, 2014
#3
+80996
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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 )

CPhill  Dec 12, 2014
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Good one :)

Phil

flflvm97  Dec 12, 2014
#5
+80996
+10

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 !!!)

CPhill  Dec 13, 2014

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