Originally Posted by

**AbYz** Hi there,

I have this : The line y = x + a , where* ***a** is a constant is a tangent to the circle (x-4)^{2} + (y-1)^{2} = 16. Find two possible values of *a*.

So am i right doing it like this:

(x-4)^{2} + (x+a-1)^{2} = 16

x^{2 }- 8x + 16 + x^{2 }+ a^{2 }+ 1 + 2xa - 2x - 2a - 16 = 0

2x^{2 }- 10x + a^{2 }+ 2xa - 2a +1 = 0

now using b^{2}-4ac:

(-10 + 2a)^{2} - 4 (2) (a^{2} - 2a + 1) which gives me

-4 (a^{2} + 6a - 23)=0

You are correct up to here. Seems you have made a mistake using the quad formula. I got -8.66 or 2.66. Try it again from here.

and that gives me possible values of 23.65 and and 12.34 when i should be getting -8.66 or 2.66.

Can anyone tell me what im doing wrong ?