HelloZello
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#1
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#1
https://gyazo.com/d1ae66a4fcc1e24e442511c7a06597d7
please help
i got up to finding the distances but am just confused on what to do right after
Last edited by HelloZello; 1 month ago
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mqb2766
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#2
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Have you sketched it?
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HelloZello
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#3
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yeah I just drew a diagram with a tangent intersecting at A and another tangent intersecting at B- I found the radius to be root 20 and CP(centre to point P(x,y) in the question said it was root 65 so I was thinking of doing pythagoras and I got AP=root 45 and next I was thinking of finding P through using the distance formula between 2 points A and P which I already know of A and using the Length of AP I got root 45=(root (6-x) squared+(-1-y)squared)-all varables are in the roots and I was thinking of finding the gradient of Ac and using the perpendicular gradient to find the equation of line intersecting at A and making y the subject and substituting in that formula I got and I'm kinda confused on what to do next since there seems to be so much working out
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HelloZello
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#4
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https://gyazo.com/9606f02d4992d072d8e153f6b133f9f0
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HelloZello
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#5
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wait I think I found my mistake hold up I didn't read the question tangents at A AND b intersect at P yikes i just followed my teachers working out
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mqb2766
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#6
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(Original post by HelloZello)
yeah I just drew a diagram with a tangent intersecting at A and another tangent intersecting at B- I found the radius to be root 20 and CP(centre to point P(x,y) in the question said it was root 65 so I was thinking of doing pythagoras and I got AP=root 45 and next I was thinking of finding P through using the distance formula between 2 points A and P which I already know of A and using the Length of AP I got root 45=(root (6-x) squared+(-1-y)squared)-all varables are in the roots and I was thinking of finding the gradient of Ac and using the perpendicular gradient to find the equation of line intersecting at A and making y the subject and substituting in that formula I got and I'm kinda confused on what to do next since there seems to be so much working out
That's the question again?
A bit of simple trig should see you good with the right diagram.
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HelloZello
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#7
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yeah-oh I never considered using trig for some reason so that seems to be making the question harder hmmmmmmm
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mqb2766
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(Original post by HelloZello)
yeah-oh I never considered using trig for some reason so that seems to be making the question harder hmmmmmmm
Can you upload what you've done / tried. For trig, read congruent triangles.
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HelloZello
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#9
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i dont have a phone at the moment srry only a pc but if I do get a solution ill try to type it here...
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mqb2766
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(Original post by HelloZello)
i dont have a phone at the moment srry only a pc but if I do get a solution ill try to type it here...
Sure, you get the equation of the circle. And for each {P,B} pair, you have two congruent right angled triangles PAC and PBC where C is the centre. That's about it, shouldn't need too much work.
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HelloZello
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#11
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(Original post by mqb2766)
Sure, you get the equation of the circle. And for each {P,B} pair, you have two congruent right angled triangles PAC and PBC where C is the centre. That's about it, shouldn't need too much work.
still not sure
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HelloZello
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#12
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nevermind
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mqb2766
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(Original post by HelloZello)
still not sure
It really would help to see a sketch of what you think the problem looks like and what you've done. You can treat it as a line intersection, ... problem, but a bit of geometry will get you there quicker.
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HelloZello
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#14
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https://gyazo.com/7e56110aab2a0cb3cf64a4054f5065dc thats the answers its just substitution after substitution it'll make you go crazy
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mqb2766
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#15
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(Original post by HelloZello)
https://gyazo.com/7e56110aab2a0cb3cf64a4054f5065dc thats the answers its just substitution after substitution it'll make you go crazy
I agree. The diagram is the most important part - the two congruent right angled triangle with lengths.
What are you asking about, given that you have the "solution"
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HelloZello
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#16
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ah no i didnt solve it- that was the answers on that question i tried to solve it without looking at the questions lol

(Original post by mqb2766)
I agree. The diagram is the most important part - the two congruent right angled triangle with lengths.
What are you asking about, given that you have the "solution"
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mqb2766
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#17
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(Original post by HelloZello)
ah no i didnt solve it- that was the answers on that question i tried to solve it without looking at the questions lol
If you have a question about what you've done, just upload it. Otherwise, if it's about the solution, what don't you understand?
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HelloZello
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#18
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-i cant upload if I don't have a phone so yh wish I did
-I think I might be confusing myself with the congruency portion of the question I tried to find the arguments and maybe use the sine rule(a/sinA=b/sinB) for finding out the length I think am overcomplicating this question,srry for being a burden
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mqb2766
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#19
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(Original post by HelloZello)
-i cant upload if I don't have a phone so yh wish I did
-I think I might be confusing myself with the congruency portion of the question I tried to find the arguments and maybe use the sine rule(a/sinA=b/sinB) for finding out the length I think am overcomplicating this question,srry for being a burden
Upload tomorrow when you have a phone?
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#20
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i dont have a phone in general- my parents have one though and it might be awkward
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