Hi,

How do i attempt, to answer part c of this question, conceptually i dont understand.

thanks

How do i attempt, to answer part c of this question, conceptually i dont understand.

thanks

Since the function is y=2x+k and it forms another tangent on C (the same gradient and parallel lines), the new point is a translation from point P and therefore after the linear vector translation from P to the center, repeat from the center to the new point and plug the x and y coordinates of the new point into the function y=2x+k where k should be -19 (Edited. I apologise for the typo error.)

(edited 2 months ago)

Original post by π/2=Σk!/(2k+1)!!

Since the function is y=2x+k and it forms another tangent on C (the same gradient and parallel lines), the new point is a translation from point P and therefore after the linear vector translation from P to the center, repeat from the center to the new point and plug the x and y coordinates of the new point into the function y=2x+k where k should be -19 (Edited. I apologise for the typo error.)

thank you. i get it now dont know why it wasnt clicking before

Thank You. Do you want to be friends? It's quite ashamedly that I say this but I don't have any friends at school (only acquaintances) and I'm in Y10 (If you want to know).

Alternatively, you could go through the whole standard motion of substituting the equation of the line into that of the circle, then check the discriminant of the resulting quadratic equation (we want discriminant=0 here since tangent=>touches at one point=> "one repeated root"), then solve for k (you'll get two - one of them should be 1 which gives the line l).

I believe that's the method most textbooks would use.

This alternative method definitely works, because the translation argument doesn't work in, for instance, find m such that y=mx is tangent to the circle. However, this is also a rites of passage for super messy algebra, so uh... just do it is the motto.

I believe that's the method most textbooks would use.

This alternative method definitely works, because the translation argument doesn't work in, for instance, find m such that y=mx is tangent to the circle. However, this is also a rites of passage for super messy algebra, so uh... just do it is the motto.

(edited 2 months ago)

Original post by π/2=Σk!/(2k+1)!!

Thank You. Do you want to be friends? It's quite ashamedly that I say this but I don't have any friends at school (only acquaintances) and I'm in Y10 (If you want to know).

Hi, as kind as that is, the age gap will be a bit weird, i hope you understand. sorry

Original post by tonyiptony

Alternatively, you could go through the whole standard motion of substituting the equation of the line into that of the circle, then check the discriminant of the resulting quadratic equation (we want discriminant=0 here since tangent=>touches at one point=> "one repeated root"), then solve for k (you'll get two - one of them should be 1 which gives the line l).

I believe that's the method most textbooks would use.

This alternative method definitely works, because the translation argument doesn't work in, for instance, find m such that y=mx is tangent to the circle. However, this is also a rites of passage for super messy algebra, so uh... just do it is the motto.

I believe that's the method most textbooks would use.

This alternative method definitely works, because the translation argument doesn't work in, for instance, find m such that y=mx is tangent to the circle. However, this is also a rites of passage for super messy algebra, so uh... just do it is the motto.

Alright, i'll try this aswell, thanks

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