# Alevel maths equations of lines

WatchPage 1 of 1

Go to first unread

Skip to page:

So I got a tricky question that i don't know how to answer can you help me?

The line with equation mx-y-2=0 touches the circle with equation x^2+6x+y^2-8y=4.

Find the two possible values of m

The line with equation mx-y-2=0 touches the circle with equation x^2+6x+y^2-8y=4.

Find the two possible values of m

1

reply

(Original post by

Try making y the subject in the equation of the line...

**tnstrk**)Try making y the subject in the equation of the line...

0

reply

Report

#4

(Original post by

I tried that and I ended up with x^2+6x+m^2x^2-12mx+16=0 and I didn't know where to go after that because there's two unkowns

**Beechey23**)I tried that and I ended up with x^2+6x+m^2x^2-12mx+16=0 and I didn't know where to go after that because there's two unkowns

So write the quadratic in terms of x, then use the quadratic formula and find the values of m where the discriminant is zero.

Last edited by mqb2766; 1 month ago

0

reply

(Original post by

You want the value(s) of m where there is a single solution to the quadratic equation in x. If there are no solutions, the line does not cut the circle. If there are two solutions, the line cuts the circle. If there is one solution, the line is a tangent to the circle.

So write the quadratic in terms of x, then use the quadratic formula and find the values of m where the discriminant is zero.

**mqb2766**)You want the value(s) of m where there is a single solution to the quadratic equation in x. If there are no solutions, the line does not cut the circle. If there are two solutions, the line cuts the circle. If there is one solution, the line is a tangent to the circle.

So write the quadratic in terms of x, then use the quadratic formula and find the values of m where the discriminant is zero.

0

reply

Report

#6

(Original post by

Sorry but I don't really understand what you mean by that

**Beechey23**)Sorry but I don't really understand what you mean by that

* miss the circle,

* graze the circumference (tangent)

* cut the circle and intersect the circumference in two places

For the quadratic in x you have, this corresponds to 0, 1 and 2 solutions. A quadratic has 1 solution (repeated) when the discriminant is zero. So write down the discriminant and set it equal to zero to find the values of m which give lines which are tangent to the circle.

Last edited by mqb2766; 1 month ago

0

reply

(Original post by

Just sketch the differerent scenarios. The line could

* miss the circle,

* graze the circumference (tangent)

* cut the circle and intersect the circumference in two places

For the quadratic in x you have, this corresponds to 0, 1 and 2 solutions. A quadratic has 1 solution (repeated) when the discriminant is zero. So write down the discriminant and set it equal to zero to find the values of m which give lines which are tangent to the circle.

**mqb2766**)Just sketch the differerent scenarios. The line could

* miss the circle,

* graze the circumference (tangent)

* cut the circle and intersect the circumference in two places

For the quadratic in x you have, this corresponds to 0, 1 and 2 solutions. A quadratic has 1 solution (repeated) when the discriminant is zero. So write down the discriminant and set it equal to zero to find the values of m which give lines which are tangent to the circle.

0

reply

Report

#8

**Beechey23**)

I tried that and I ended up with x^2+6x+m^2x^2-12mx+16=0 and I didn't know where to go after that because there's two unkowns

0

reply

(Original post by

You may wish to group like terms..

**tnstrk**)You may wish to group like terms..

0

reply

Report

#10

(Original post by

It's the part about putting x^2+6x+m^2x^2-12mx+16=0 into a quadratic that's Confusing me because I dont see how you would do it

**Beechey23**)It's the part about putting x^2+6x+m^2x^2-12mx+16=0 into a quadratic that's Confusing me because I dont see how you would do it

x^2*() + x*() + () =

0

reply

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top