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MATHS QUESTION HELP: Core 2

I can do 19 (a), I just need help on 19 (b) and (c)! Thank you :smile:

Reply 1
Original post by creativebuzz
I can do 19 (a), I just need help on 19 (b) and (c)! Thank you :smile:



For (b), if you know the centre O of the circle, think about a radius OP joining the centre to the point.

What's the gradient of this radius line in terms of x, y and the coordinates of the circle?

How does the gradient of the tangent at P relate to the gradient of the radius OP?
Reply 2
Original post by creativebuzz
I can do 19 (a), I just need help on 19 (b) and (c)! Thank you :smile:



for b)

differentiate w.r.t x the equation of the circle with implicit differentiation
and arrange to y'

for c)

the equation of the tangent line in (x0, y0) point of the circe
yy0=m(xx0)y-y_0=m(x-x_0)

when m=y'(x0,y0)
Reply 3
Original post by ztibor
for b)

differentiate w.r.t x the equation of the circle with implicit differentiation
and arrange to y'

for c)

the equation of the tangent line in (x0, y0) point of the circe
yy0=m(xx0)y-y_0=m(x-x_0)

when m=y'(x0,y0)


This is the C2 module - the OP is unlikely to have seen implicit differentiation before :smile:
Original post by davros
For (b), if you know the centre O of the circle, think about a radius OP joining the centre to the point.

What's the gradient of this radius line in terms of x, y and the coordinates of the circle?

How does the gradient of the tangent at P relate to the gradient of the radius OP?


For b) I got 4-x/y-8

c) y - 8 = 4-x/y-8 (x-21)

y^2 - 16y + 64 = -x^2 + 25x -84

Where did I got wrong? :/
Reply 5
Original post by creativebuzz
For b) I got 4-x/y-8

c) y - 8 = 4-x/y-8 (x-21)

y^2 - 16y + 64 = -x^2 + 25x -84

Where did I got wrong? :/


You're making this much too difficult for yourself!

The gradient of the tangent is (4-x)/(y-8) as you say. At the point (21, 8) you have y-8 = 0 so the gradient of the tangent is undefined. So what sort of line is the tangent at that point? So what is the equation of the tangent at that point?
Original post by davros
You're making this much too difficult for yourself!

The gradient of the tangent is (4-x)/(y-8) as you say. At the point (21, 8) you have y-8 = 0 so the gradient of the tangent is undefined. So what sort of line is the tangent at that point? So what is the equation of the tangent at that point?


Why is y-8=0?
Reply 7
Original post by creativebuzz
Why is y-8=0?


At the point (21, 8), x = 21 and y = 8. So y - 8 = 0 :smile:
Original post by davros
At the point (21, 8), x = 21 and y = 8. So y - 8 = 0 :smile:


Oh yeah I understand that, I was wondering what equation you subbed that into to get y - 8 = 0? :smile:

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