Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Core 1, Question

Im stuck with this question can someone help please?

A line has equation y=2k-x where k is a constant

show that the x coordiante of any point of intersection of the line and the circle satfisies the equation

x^2-2(k+1)x+2k^2-7=0

You didn't type the equation of the circle.

You need to provide us with the equation of the circle in order to prove it satisfies the equation.

However, what you would do is solve the equations simultaneously/use the substitution method - always do this whenever it says 'intersection'.

(edited 12 years ago)

OP

sorry, the equation of the circle is x^2+y^2-4x-14=0

OP

okk nvm i got it now, but the next question is find the values of k which the equation x^2-2(k+1)x+2k^2-7=0 has equal roots, so we substiute it into the discriminant but im not too sure which is the value b

Original post by Gary

okk nvm i got it now, but the next question is find the values of k which the equation x^2-2(k+1)x+2k^2-7=0 has equal roots, so we substiute it into the discriminant but im not too sure which is the value b

Okay so you have a quadratic:

x^2-2(k+1)x+2k^2-7= Ax^2+ bx+c

A= 1

B= -2(k+1)

c= 2k^2 -7

Then apply the correct discriminant.

OP

Yeahh thats what i thought aswell but the asnwer said b is 4(k+1)^2 instead of -2(k+1)^2 so im a bit confused

Original post by Gary

Yeahh thats what i thought aswell but the asnwer said b is 4(k+1)^2 instead of -2(k+1)^2 so im a bit confused

Yes because its (-2(k+1))^2 so (-2)^2= 4 Yes?

-> 4(k+1)^2

OP

Ohh right i didn't know you had to square the -2 aswell but thank you :smile:

:smile:

Original post by Gary

Ohh right i didn't know you had to square the -2 aswell but thank you :smile:

:smile:

Simple indicies rule: (ab)^2= a^2b^2.

OP

okk i expandd it and got 4k^2 +8k + 8 and for the -4(2k^2-7) i got -8k^2+28

i think ive done something wrong here, correct me please?

Original post by Gary

okk i expandd it and got 4k^2 +8k + 8 and for the -4(2k^2-7) i got -8k^2+28

i think ive done something wrong here, correct me please?

You have a +8 for some odd reason.

4(k+1)^2= 4(k^2+2k+1)= 4k^2+8k+4

The second bit is right though.

4(k+1)^2=4(k^2+2k+1)=4k^2+8k+4

Your expansion of -4(2k^2-7) is correct.

OP

Ohh right i know where ive gone wrong now, did some silly mistakes. But thanks for all your help :smile:

:smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you

Your guide to the ultimate essay

Your guide to the ultimate essay

A guide to free resources for GCSE and A-level exam revision

A guide to free resources for GCSE and A-level exam revision

How to write an excellent personal statement in 10 steps

How to write an excellent personal statement in 10 steps

I want to go to uni but I don't know what to study

I want to go to uni but I don't know what to study