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Maths Question

Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.
The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C. Work out coordinates of C
I’ll put pic in reply section too
I don’t know how to do it
Reply 2
Original post by Defence11
Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.
The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C. Work out coordinates of C
I’ll put pic in reply section too
I don’t know how to do it


There is some information missing. Can you post a picture of the full question?
You need to add (a) the equation and (b) the x-coordinate of B. Then we can help you :wink:
Reply 4
Original post by mqb2766
There is some information missing. Can you post a picture of the full question?

i’ve already mentioned the info u need in my post
Reply 5
Original post by Justvisited
You need to add (a) the equation and (b) the x-coordinate of B. Then we can help you :wink:

i don’t understand sorry. could u go through the steps pls
Original post by Defence11
i don’t understand sorry. could u go through the steps pls

Re-read your OP:

"Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.
The point B also lies on the circle and has an x-coordinate of ."

(1) Where?
(2) Strong syntax.
Reply 7
Original post by mqb2766
There is some information missing. Can you post a picture of the full question?

Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.

The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C.
Reply 8
Original post by Justvisited
Re-read your OP:

"Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.
The point B also lies on the circle and has an x-coordinate of ."

(1) Where?
(2) Strong syntax.

Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.

The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C.
Reply 9
Original post by Defence11
Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.

The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C.


What is the x-coordinate of B .... ?
Can you pls post a picture of the question?
Original post by Defence11
Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.

The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C.

Please stop taking the mick.
Here's the question in full.
Original post by BuryMathsTutor
Here's the question in full.

Thank you, and well spotted!

Without giving a complete solution, the basic steps are as follows:

(1) Use the equation to calculate the y-coordinate at A.
(2) Use the quotient of the y and x coordinates at A to work out the gradient of the radius OA.
(3) Take the negative reciprocal of (2) to get the slope of the tangent at A. Call this m.
(4) Now use the general equation of a straight line with the above values of y, m and x, to work out the c in the equation and thus complete it.
(5) Repeat steps (1) to (4) with point B (working out the x-coordinate from y instead of vice-versa).
(6) You now have two linear equations in x and y which you can solve as you normally solve simultaneous equations, to obtain the coordinates of their point of intersection.

OP, please come back if you're having trouble with any of the above.
try implicit differentiation so that you can get the equation of the 2 tangents. Or if you don't know that find dy/dx by rearranging the equation.

you should be able to work out what else is required to find the answer...
(edited 6 years ago)
Original post by TheTroll73
try implicit differentiation so that you can get the equation of the 2 tangents. Or if you don't know that find dy/dx by rearranging the equation.

you should be able to work out what else is required to find the answer...

That works of course, but this is definitely a pre-calc, GCSE year type of question. Your method will cover all kinds of curves, but in the unique case of a circle one is supposed to take advantage of the fact that the tangent is always perpendicular to a line from the origin to that point (which for circles is called the 'radius').
Original post by Justvisited
That works of course, but this is definitely a pre-calc, GCSE year type of question. Your method will cover all kinds of curves, but in the unique case of a circle one is supposed to take advantage of the fact that the tangent is always perpendicular to a line from the origin to that point (which for circles is called the 'radius':wink:.


thought it was an AS math question since I don't remember being told about the equation of a circle in GCSE...
use
1. product of two lines perpendicular to each other is -1
2. angle between radius and tangent to the circle is 90

A is (-2,-4) and B is (sqrt 10, sqrt 10)
let tangent line from a l1 and from b l2

then l1 :
the gradient of radius from circle centre to a has (0--4)/(0--2)=2
from 1. the gradient of l1 is -0.5
therefore l1 : y=-0.5(x+2)-4=-0.5x-5

l2:
gradeint of radius from O to b has (sqrt 10)/(sqrt 10)=1
from 1. gradient of l2 is -1.
therefore l2: y= -(x-sqrt10)+sqrt10 = -x + 2sqrt10

C is where the 2 tangents meet so
-0.5x - 5 = -x + 2sqrt10
0.5x = 2sqrt10 + 5
x= 4sqrt10 + 10
sub x value into any line bc its on both lines
y=-0.5(4sqrt10+10)-5 = -2sqrt10-10

c(4sqrt10 + 10 , -2sqrt10-10)

graph is aut ism could be drawn for case A x coordinate is -4 and y coordinate is -2
but could simply know its wrong by the slope of 2 lines.

illusion 100
(edited 6 years ago)
Original post by TheTroll73
thought it was an AS math question since I don't remember being told about the equation of a circle in GCSE...

I think the new 1-9 GCSE may be tougher and include such questions, without moving into calculus. But even at AS, this kind of Q would definitely be put without expecting people to know implicit differentiation which is a Y13 topic.
Reply 18
Original post by username4112574
use
1. product of two lines perpendicular to each other is -1
2. angle between radius and tangent to the circle is 90
A is (-2,-4) and B is (sqrt 10, sqrt 10)
let tangent line from a l1 and from b l2
then l1 :
the gradient of radius from circle centre to a has (0--4)/(0--2)=2
from 1. the gradient of l1 is -0.5
therefore l1 : y=-0.5(x+2)-4=-0.5x-5
l2:
gradeint of radius from O to b has (sqrt 10)/(sqrt 10)=1
from 1. gradient of l2 is -1.
therefore l2: y= -(x-sqrt10)+sqrt10 = -x + 2sqrt10
C is where the 2 tangents meet so
-0.5x - 5 = -x + 2sqrt10
0.5x = 2sqrt10 + 5
x= 4sqrt10 + 10
sub x value into any line bc its on both lines
y=-0.5(4sqrt10+10)-5 = -2sqrt10-10
c(4sqrt10 + 10 , -2sqrt10-10)
graph is aut ism could be drawn for case A x coordinate is -4 and y coordinate is -2
but could simply know its wrong by the slope of 2 lines.
illusion 100

Isn't that wrong since the x value has to be negative while the y value has to be positive because C is in the 1st quadrant on the left?
Original post by Defence11
Consider the circle with equation sketched below. The point A lies on the circle and has a y-coordinate of -4.
The point B also lies on the circle and has an x-coordinate of . The tangent line at A intersects the tangent line at B at point C. Work out coordinates of C
I’ll put pic in reply section too
I don’t know how to do it

"The tangent at any point of a circle is perpendicular to the radius through the point of contact"
so we know that if two linear are perpendicular to each other if we multiply both slopes it'll always be -1
and we know that this circle center is 0,0 if we find slop from (0,0) to (√10,√10) it'll be 1 and from (0,0) to (2,-4 ) it'll be -2 so we know that linear's slop are -1,1/2 in order
from linear formular we know that y=mx+b if x=√10 and y=√10 and m=-1 blah blah blah we got b=2√10
and the same thing as the second linear we got b=-5
we know that if two line touch each other we write
(-1)x+2√10= 1/2x-5
we got x and find y next

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