Circle C has a center of (−3, 1) and a radius of 4. Circle E has a center of (4, 9) and a radius of 3. Prove that the two circles are similar. Anyone please help me with this? @amistre64 @Hero @ash2326 @Luigi0210
@Whitemonsterbunny17
what is your definition of similarity?
i can see it as side-angle-side ...
but thats congruent .... similar is any scalar variety
move each circle to the center, and scale one to get the other .... but like a stated first, depends on your definition of similar
well i went to see how to do this on geogebra as it says in my lesson to create a circle with a center of the given coordinates and whichever radius so i did that with both the circles and tried to translate it to lie on the same center point of circle B but i didn't know ho to. @amistre64
thats why i just move them to the center ... if we are allowed to move one object, we are allowed to move both objects ...
x^2 + y^2 = 4^2 and x^2 + y^2 = 3^2 the one is just a scalar of the other by 4^2/3^2
so thats how i prove that the circles are similar?
thats how I would do it, but im not the one doing a specifically instructed assignment
the assignment just gave us the center coordinate points and their radiuses and said to prove that the 2 circles are similar. @amistre64
@amistre64
@Hero
@amistre64
@amistre64
@agent0smith
All circles are similar...
yeah prove it is what the question is asking.. i know that all circles are similar.
translating something is just moving it about the grid ... so if we 'move' both centers the the origin (0,0) we have moved one on top of the other ....
can u show me how on geogebra cuz i tried its not working...
i dont know how to work geogebra ...
but the concept i have in mind is this: |dw:1404157014270:dw| take some random point on one of the circles, we form a triangle with it .... which can then be scaled to find a spot for it on the other circle.
since x = rcos(t) and y = r sin(t) some (a,b) on the 3rad circle is just: a = 3cos(t), b = 3sin(t) multiply by 4/3 to scale it to the 4rad circle 4a/3 = 4cos(t), 4b/3 = 4sin(t) now a' = 4a/3 and b' = 4b/3 for the point that is similar on the other circle.
but im not sure how geogebra works to use this idea
this is what it says on my lesson
and thats what i demonstrated, except i moved both to center at 0,0 indtead
*instead
the scale factor ak = a' is when k=4/3 tom scale from 3 to 4
x^2 + y^2 = 3^2 , for some point (a,b) 4^2/3^2 * (a^2 + b^2 = 3^2) (4a/3)^2 + (4b/3)^2 = 4^2 therefore, by using a dilation factor of k: ak = a' 4a/3 = a', k=4/3
but how does that prove theyre similar?
thats why i said it depends on your definition of similar ... your file suggests that the definition is that a dilated objects are similar to each other.
our dilation factor to get from a to a' is 4/3 and to get from b to b' is 4/3; so we have demonstarted that we can get from one point (a,b) on one circle to a point (a',b') on the other circle by a dilation of 4/3
ok but for the assignment i have to dilate circle A to lie on the center point of circle B so when i dilate it on geogebra and it asks for the factor should i just type 4/3 in?
maybe, but i dont know how to use geogebra so asking me how to do somehting with geogebra is futile.
a center of dilation is key ..... which is why i simply made both centers at the origin
yeah thats my problem now.. my lesson has the example using geogebra and when i tried using it i had no idea how to.
:) i have to go, kids need a ride .... good luck with the geogebra specifics tho.
thanks..
@agent0smith can u please help?
try inputing 4/3 and see if it works is the best adivise ive got
@agent0smith
@dan815 plz help
No idea.
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