# circles help(hardcore question)

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https://gyazo.com/d1ae66a4fcc1e24e442511c7a06597d7

please help

i got up to finding the distances but am just confused on what to do right after

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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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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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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#6

(Original post by

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

**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

A bit of simple trig should see you good with the right diagram.

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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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#8

(Original post by

yeah-oh I never considered using trig for some reason so that seems to be making the question harder hmmmmmmm

**HelloZello**)yeah-oh I never considered using trig for some reason so that seems to be making the question harder hmmmmmmm

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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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#10

(Original post by

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...

**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...

Last edited by mqb2766; 1 month ago

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(Original post by

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.

**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.

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#13

(Original post by

still not sure

**HelloZello**)still not sure

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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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#15

(Original post by

https://gyazo.com/7e56110aab2a0cb3cf64a4054f5065dc thats the answers its just substitution after substitution it'll make you go crazy

**HelloZello**)https://gyazo.com/7e56110aab2a0cb3cf64a4054f5065dc thats the answers its just substitution after substitution it'll make you go crazy

What are you asking about, given that you have the "solution"

Last edited by mqb2766; 1 month ago

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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

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"

**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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#17

(Original post by

ah no i didnt solve it- that was the answers on that question i tried to solve it without looking at the questions lol

**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

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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

-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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#19

(Original post by

-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

**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

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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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