Here is the question that I am stuck on:
Find the equation of the tangents drawn from the point (4,3) to the circle x^2 + y^2 = 5
I know that the aim is to find the gradients of the tangents, thus enabling use of y  y1 = m(x  x1). However I have been struggling for a while to make any progress on this.
Thanks in advance for any help.

Sm0key
 Follow
 0 followers
 1 badge
 Send a private message to Sm0key
 Thread Starter
Offline1ReputationRep: Follow
 1
 15052015 23:50

lizard54142
 Follow
 7 followers
 3 badges
 Send a private message to lizard54142
Offline3ReputationRep: Follow
 2
 15052015 23:53
(Original post by Sm0key)
Here is the question that I am stuck on:
Find the equation of the tangents drawn from the point (4,3) to the circle x^2 + y^2 = 5
I know that the aim is to find the gradients of the tangents, thus enabling use of y  y1 = m(x  x1). However I have been struggling for a while to make any progress on this.
Thanks in advance for any help. 
 Follow
 3
 15052015 23:54
(Original post by Sm0key)
Here is the question that I am stuck on:
Find the equation of the tangents drawn from the point (4,3) to the circle x^2 + y^2 = 5
I know that the aim is to find the gradients of the tangents, thus enabling use of y  y1 = m(x  x1). However I have been struggling for a while to make any progress on this.
Thanks in advance for any help.
Posted from TSR Mobile 
Sm0key
 Follow
 0 followers
 1 badge
 Send a private message to Sm0key
 Thread Starter
Offline1ReputationRep: Follow
 4
 16052015 00:12
(Original post by lizard54142)
Have you covered implicit differentiation? This would be the easiest way to get the gradient, or you could rearrange into the form y = ... 
XI Ki11JoY IX
 Follow
 0 followers
 1 badge
 Send a private message to XI Ki11JoY IX
Offline1ReputationRep: Follow
 5
 16052015 00:37
(Original post by Sm0key)
Here is the question that I am stuck on:
Find the equation of the tangents drawn from the point (4,3) to the circle x^2 + y^2 = 5
I know that the aim is to find the gradients of the tangents, thus enabling use of y  y1 = m(x  x1). However I have been struggling for a while to make any progress on this.
Thanks in advance for any help.
Well, first of all, the question has a few typos, the equation for a circle (with centre 0,0 is) x^2 + y^2 = r^2 , a quick check of the coordinates can also prove that r^2 is in fact 25.
Second, there is only 1 tangent at a point, so that might help.
right, y^2 = 25  x^2
implicit differentiation gives 2y(dy/dx) = 2x
dy/dx = =2x/2y or x/y or 4/3
y= 4/3x + k
k = 8
Tangent: y=4/3x 8
In from ax+bx+c=0
4x3y24=0
Edit: I'm dumb, sorry for being a jag, I'll insert another comment explaining what you are supposed to do.Last edited by XI Ki11JoY IX; 16052015 at 16:29. 
Sm0key
 Follow
 0 followers
 1 badge
 Send a private message to Sm0key
 Thread Starter
Offline1ReputationRep: Follow
 6
 16052015 02:44
(Original post by XI Ki11JoY IX)
Circle geometry is the best.
Well, first of all, the question has a few typos, the equation for a circle (with centre 0,0 is) x^2 + y^2 = r^2 , a quick check of the coordinates can also prove that r^2 is in fact 25.
Second, there is only 1 tangent at a point, so that might help.
right, y^2 = 25  x^2
implicit differentiation gives 2y(dy/dx) = 2x
dy/dx = =2x/2y or x/y or 4/3
y= 4/3x + k
k = 8
Tangent: y=4/3x 8
In from ax+bx+c=0
4x3y24=0
ie radius = sqrt(5). 
physicsmaths
 Follow
 146 followers
 18 badges
 Send a private message to physicsmaths
Offline18ReputationRep: Follow
 7
 16052015 07:25
(Original post by Sm0key)
The question on the paper is definitely: "tangents from (4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5).
There would be two tangents that would go this way.
Posted from TSR Mobile 
BuryMathsTutor
 Follow
 8 followers
 13 badges
 Send a private message to BuryMathsTutor
 Visit BuryMathsTutor's homepage!
Offline13ReputationRep: Follow
 8
 16052015 07:47
Any straight line through (4,3) has equation or .
You can substitute for y in the equation of the circle to get a quadratic in x^2. Now think about the discriminant. This will allow you to find the two possible gradients.
There's probably an easier way but I'm still half asleep...
Last edited by BuryMathsTutor; 16052015 at 07:53. Reason: Added solutions 
 Follow
 9
 16052015 07:52
How many marks?

physicsmaths
 Follow
 146 followers
 18 badges
 Send a private message to physicsmaths
Offline18ReputationRep: Follow
 10
 16052015 08:11
(Original post by XI Ki11JoY IX)
Circle geometry is the best.
Well, first of all, the question has a few typos, the equation for a circle (with centre 0,0 is) x^2 + y^2 = r^2 , a quick check of the coordinates can also prove that r^2 is in fact 25.
Second, there is only 1 tangent at a point, so that might help.
right, y^2 = 25  x^2
implicit differentiation gives 2y(dy/dx) = 2x
dy/dx = =2x/2y or x/y or 4/3
y= 4/3x + k
k = 8
Tangent: y=4/3x 8
In from ax+bx+c=0
4x3y24=0
Posted from TSR Mobile 
BuryMathsTutor
 Follow
 8 followers
 13 badges
 Send a private message to BuryMathsTutor
 Visit BuryMathsTutor's homepage!
Offline13ReputationRep: Follow
 11
 16052015 08:31

lizard54142
 Follow
 7 followers
 3 badges
 Send a private message to lizard54142
Offline3ReputationRep: Follow
 12
 16052015 08:50
(Original post by Sm0key)
Yes I have. I'm afraid I don't follow how that would be of use here? I mean: what equation must I form to then implicitly differentiate? 
 Follow
 13
 16052015 08:50
(Original post by Sm0key)
The question on the paper is definitely: "tangents from (4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5). 
XI Ki11JoY IX
 Follow
 0 followers
 1 badge
 Send a private message to XI Ki11JoY IX
Offline1ReputationRep: Follow
 14
 16052015 16:27
(Original post by Sm0key)
The question on the paper is definitely: "tangents from (4,3) to the circle x^2 + y^2 = 5"
ie radius = sqrt(5). 
physicsmaths
 Follow
 146 followers
 18 badges
 Send a private message to physicsmaths
Offline18ReputationRep: Follow
 15
 16052015 16:33
(Original post by XI Ki11JoY IX)
That's impossible.
Posted from TSR Mobile 
XI Ki11JoY IX
 Follow
 0 followers
 1 badge
 Send a private message to XI Ki11JoY IX
Offline1ReputationRep: Follow
 16
 16052015 16:56

Sm0key
 Follow
 0 followers
 1 badge
 Send a private message to Sm0key
 Thread Starter
Offline1ReputationRep: Follow
 17
 16052015 17:03
Q9 circle & 2 tangents.pdf
It seems that some members don't trust my typing of the question  for everyone's benefit here is a scan of the question, with my sketch too.
Edit: After implicit diff. of the circle equation I get:
dy/dx = x/y
But I can't see what 2 pairs of (x,y) values I need (obviously lying on the circumference) to be able get the two tangents' gradients?Last edited by Sm0key; 16052015 at 17:08. 
BuryMathsTutor
 Follow
 8 followers
 13 badges
 Send a private message to BuryMathsTutor
 Visit BuryMathsTutor's homepage!
Offline13ReputationRep: Follow
 18
 16052015 17:59
(Original post by Sm0key)
Q9 circle & 2 tangents.pdf
It seems that some members don't trust my typing of the question  for everyone's benefit here is a scan of the question, with my sketch too.
Edit: After implicit diff. of the circle equation I get:
dy/dx = x/y
But I can't see what 2 pairs of (x,y) values I need (obviously lying on the circumference) to be able get the two tangents' gradients?
See posts 8 and 11. Both hints I've given will get you to the answer quite quickly. Post again if you need more. 
Sm0key
 Follow
 0 followers
 1 badge
 Send a private message to Sm0key
 Thread Starter
Offline1ReputationRep: Follow
 19
 17052015 01:50
(Original post by BuryMathsTutor)
See posts 8 and 11. Both hints I've given will get you to the answer quite quickly. Post again if you need more.
I've ended up with:
(1 + m^2)x^2 + (8m^2 + 6m)x + (16m^2 + 24m  4) = 0
following the substitution. However I cannot decide what requirement I must place on the discriminant. I imagine that b^2 > 4ac is required to give 2 real solutions. However I can't work out how the above quadratic would handle 2 *tangents* as opposed to 2 lines with 2 intercepts each (ie 4 solutions). How do I know that consideration of the discriminant will produce only 2 solutions (corresponding to tangents)? 
BuryMathsTutor
 Follow
 8 followers
 13 badges
 Send a private message to BuryMathsTutor
 Visit BuryMathsTutor's homepage!
Offline13ReputationRep: Follow
 20
 17052015 08:06
(Original post by Sm0key)
Thanks for the pointers.
I've ended up with:
(1 + m^2)x^2 + (8m^2 + 6m)x + (16m^2 + 24m  4) = 0
following the substitution. However I cannot decide what requirement I must place on the discriminant. I imagine that b^2 > 4ac is required to give 2 real solutions. However I can't work out how the above quadratic would handle 2 *tangents* as opposed to 2 lines with 2 intercepts each (ie 4 solutions). How do I know that consideration of the discriminant will produce only 2 solutions (corresponding to tangents)?
You have a small error in your equation by the way. The 4 should be +4.
You should get a quadratic in m:
or
 1
 2
Related discussions
Related university courses

Mathematics with Finance
University of Manchester

Mathematics and Computer Science
Queen's University Belfast

Mathematics and Physics
University of Surrey

Mathematics and Statistics
University of Edinburgh

Initial Year for Extended Degree in Science  Mathematics
University of Hertfordshire

Mathematics
University of Oxford

Mathematics (4 years)
Durham University

Mathematics (4 years)
Durham University

Mathematics and Statistics
University of Oxford

Mathematics with Foundation Year
University of Southampton
see more
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 Lemur14
 brainzistheword
 Quirky Object
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel
 Toastiekid
 CinnamonSmol
 Qer
 The Empire Odyssey
 RedGiant
 Sinnoh