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strange tangent question

gcse math edexcel

i know how to find the equation of a tangent from a circle and this is my method:

change in y/change in x = gradient
recipricol it
put a - infront of it
do y=mx+c and substitute value of M in from above
put the values of y and x in that you know from the point of intersection and solve for C
and rewrite equation
i cant find an image but ill try and draw it for you.




this is the question im faced with, as said im familiar with tangent stuff but the tangent here is in a different place, and also the co-ordinates of x and y are in a strange format, how do i find the equation and what do i need to differ from the method i know? thanks!
help me.PNG13.4KB
help me omg.PNG72.7KB
Original post by Josh827
gcse math edexcel

i know how to find the equation of a tangent from a circle and this is my method:

change in y/change in x = gradient
recipricol it
put a - infront of it
do y=mx+c and substitute value of M in from above
put the values of y and x in that you know from the point of intersection and solve for C
and rewrite equation
i cant find an image but ill try and draw it for you.




this is the question im faced with, as said im familiar with tangent stuff but the tangent here is in a different place, and also the co-ordinates of x and y are in a strange format, how do i find the equation and what do i need to differ from the method i know? thanks!


It's the same method. All you need to do is determine pp first because we have enough information to determine it.

A(p,−15)A(p,-15) is on the circle so p2+(−15)2=261p^2+(-15)^2=261 is a condition pp must satisfy.

Once you got A, just proceed with your usual method.
Reply 2
Avatar for Josh827
Josh827
OP
11
Original post by RDKGames
It's the same method. All you need to do is determine pp first because we have enough information to determine it.

A(p,−15)A(p,-15) is on the circle so p2+(−15)2=261p^2+(-15)^2=261 is a condition pp must satisfy.

Once you got A, just proceed with your usual method.


im confused because it says P, -15 and P is bigger than 0, to find P do i do the square root of 261, so the co-ordinates are (square root of 261, -15)? And i thought that maybe i would get rid of the - sign because the tangent isnt sloping downwards? thanks
Original post by Josh827
im confused because it says P, -15 and P is bigger than 0, to find P do i do the square root of 261, so the co-ordinates are (square root of 261, -15)? And i thought that maybe i would get rid of the - sign because the tangent isnt sloping downwards? thanks


pp is just the x-coordinate of AA - it is not the slope, therefore your reasoning is not correct.

We have two conditions which pp statisfies:

p>0p>0
p2+(−15)2=261p^2+(-15)^2=261

To find pp you need to solve the second equation (which you've done unsuccessfully) which gives two possible values for pp so you need to impose the first condition to get the value for it.
Reply 4
Avatar for SG!YK.VN
SG!YK.VN
9
Original post by Josh827
gcse math edexcel

i know how to find the equation of a tangent from a circle and this is my method:

change in y/change in x = gradient
recipricol it
put a - infront of it
do y=mx+c and substitute value of M in from above
put the values of y and x in that you know from the point of intersection and solve for C
and rewrite equation
i cant find an image but ill try and draw it for you.




this is the question im faced with, as said im familiar with tangent stuff but the tangent here is in a different place, and also the co-ordinates of x and y are in a strange format, how do i find the equation and what do i need to differ from the method i know? thanks!


For the q22, I think it's done like this:

(p,-15) lies on the circumference, therefore it should go into the circle equation x^2 + y^2 = 261

so substitute it in and it is:
p^2 + (-15)^2 =261

then you solve it.

p^2 = 261 - 225
p^2 = 36
p = 6
Reply 5
Avatar for Josh827
Josh827
OP
11
Original post by RDKGames
pp is just the x-coordinate of AA - it is not the slope, therefore your reasoning is not correct.

We have two conditions which pp statisfies:

p>0p>0
p2+(−15)2=261p^2+(-15)^2=261

To find pp you need to solve the second equation (which you've done unsuccessfully) which gives two possible values for pp so you need to impose the first condition to get the value for it.

p^2+-(15)^2=261
261-(15)^2=225
261-225=36
p^2= 36 so p=6

so A is at the point (6,-15)
-15/6 = - 2.5 (-5/2)
recipricol we get -2/5
get rid of minus sign we get 2/5
y=2/5x+c
-15=2/5(6)+c
-15=12/5 +c
-15 times -5 = c
so c= 75

y=2/5x+75

this correct?
(edited 6 years ago)
Original post by Josh827

-15=12/5 +c
-15 times -5 = c
so c= -75

y=2/5x-75

this correct?


No.

-15 times -5 is NOT c
Reply 7
Avatar for Josh827
Josh827
OP
11
Original post by RDKGames
No.

-15 times -5 is NOT c


is it correct now
Reply 8
Avatar for Josh827
Josh827
OP
11
eh i just checked the mark scheme apparently c is -17.4?
Original post by Josh827
is it correct now


Not sure what you've changed. My comment has not been addressed by the looks of it.
Reply 10
Avatar for Josh827
Josh827
OP
11
Original post by RDKGames
Not sure what you've changed. My comment has not been addressed by the looks of it.


i changed it to positive 75 since -15 x -5 == 75
Original post by Josh827
p^2+-(15)^2=261
261-(15)^2=225
261-225=36
p^2= 36 so p=6

so A is at the point (6,-15)
-15/6 = - 2.5 (-5/2)
recipricol we get -2/5
get rid of minus sign we get 2/5
y=2/5x+c
-15=2/5(6)+c
-15=12/5 +c
-15 times -5 = c
so c= 75

y=2/5x+75

this correct?


No, u went wrong in the last bit.

-15=12/5 +c
-15 -12/5 = c
-17.4 =c

y=2/5x -17.4
Original post by Josh827
i changed it to positive 75 since -15 x -5 == 75


You should've just subtracted 12/5 from both sides for cc, and shown above.
Reply 13
Avatar for Josh827
Josh827
OP
11
Original post by SG!YK.VN
No, u went wrong in the last bit.

-15=12/5 +c
-15 -12/5 = c
-17.4 =c

y=2/5x -17.4


ahh i see i rearranged it wrong, so annoying that you can ruin such a hard question with just a simple error :frown: thanks.

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