# Math Help - possible value

1. ## possible value

hi, my question is:

find the possible value of the constant c so that the line $y = 2x + c$ is a tangent to the circle $x^2 + y^2 = 4$

i treated it as a simultaneous equation (if thats the right thing to do) and continued by making equation one $y^2 = (2x + c)^2$ and subbed it into the second like this $x^2 + (2x + c)^2 = 4$ then expanded the second to $x^2 + 4x^2 + 4xc + c^2 = 4$ then simplified to $5x^2 + 4xc + c^2 = 4$

i don't know what to do from here, can someone help me out please? thanks

2. Hello mark
Originally Posted by mark
hi, my question is:

find the possible value of the constant c so that the line $y = 2x + c$ is a tangent to the circle $x^2 + y^2 = 4$

i treated it as a simultaneous equation (if thats the right thing to do) and continued by making equation one $y^2 = (2x + c)^2$ and subbed it into the second like this $x^2 + (2x + c)^2 = 4$ then expanded the second to $x^2 + 4x^2 + 4xc + c^2 = 4$ then simplified to $5x^2 + 4xc + c^2 = 4$

i don't know what to do from here, can someone help me out please? thanks
Thanks for showing us your working so far. You are quite correct in what you've done. The next step is to re-arrange as a quadratic in $x$:

$5x^2 +4c\cdot x+c^2-4=0$

and then say that the line is a tangent if the two points of intersection coincide at one point; i.e. if this quadratic has equal roots. So, use $b^2-4ac=0$, and solve for $c$.

(I make the answers $c = \pm 2\sqrt5$.)

3. thanks for answering but i don't think i understand what you mean by rearrange as a quadratic in x? why have you set it out like this $5x^2 +4c\cdot x+c^2-4=0$?

if i was using $\sqrt{b^2 - 4ac} = 0$ i would have made it $\sqrt{16x^2c^2 - 4(5)(-4)} = 0$

4. Hello mark
Originally Posted by mark
thanks for answering but i don't think i understand what you mean by rearrange as a quadratic in x? why have you set it out like this $5x^2 +4c\cdot x+c^2-4=0$?

if i was using $\sqrt{b^2 - 4ac} = 0$ i would have made it $\sqrt{16x^2c^2 - 4(5)(-4)} = 0$
I mean re-arrange to form an equation of the form $ax^2 + bx + c=0$.

This equation has equal roots if $b^2-4ac=0$.

So in the equation $5x^2 +4c\cdot x+c^2-4=0$

• $a=5$
• $b=4c$
• $c=(c^2-4)$

So $b^2 - 4ac = (4c)^2 - 4\cdot5\cdot(c^2-4) = 16c^2-20(c^2-4)=80-4c^2$

So $b^2-4ac = 0 \Rightarrow c^2=20 \Rightarrow c = \pm2\sqrt5$