1. ## tangent

show that the line $\displaystyle 8x + y + 10 = 0$ is a tangent to the circle $\displaystyle x^2 + y^2 - 6x + 3y - 5 = 0$

i started by making y the subject in the first equation so $\displaystyle y = -10 -8x = 0$ and subbed it into the second making it $\displaystyle x^2 + (-10 - 8x)^2 - 6x + 3(-10 -8x) - 5 = 0$ then simplified it to $\displaystyle x^2 + 100 + 160x + 64x^2 - 6x - 30 - 24x - 5 = 0$ then simplified it further to $\displaystyle 65x^2 + 128x + 65 = 0$. so if its tangent then $\displaystyle \sqrt {b^2 - 4ac}$ must equal 0. but using the discriminant theory it would end up as $\displaystyle \sqrt{16384 - 16900}$ which clearly isn't 0. can someone show me how its done please?

thanks

2. Originally Posted by mark
show that the line $\displaystyle 8x + y + 10 = 0$ is a tangent to the circle $\displaystyle x^2 + y^2 - 6x + 3y - 5 = 0$

i started by making y the subject in the first equation so $\displaystyle y = -10 -8x = 0$ and subbed it into the second making it $\displaystyle x^2 + (-10 - 8x)^2 - 6x + 3(-10 -8x) - 5 = 0$ then simplified it to $\displaystyle x^2 + 100 + 160x + 64x^2 - 6x - 30 - 24x - 5 = 0$ then simplified it further to

$\displaystyle 65x^2 + 128x + 65 = 0$. Mr F says: This mistake is here. See below.

so if its tangent then $\displaystyle \sqrt {b^2 - 4ac}$ must equal 0. but using the discriminant theory it would end up as $\displaystyle \sqrt{16384 - 16900}$ which clearly isn't 0. can someone show me how its done please?

thanks
The simplification should be $\displaystyle 65x^2 + 130x + 65 = 0$.
This in turn is equivalent to $\displaystyle 0 = 65 (x^2 + 2x + 1) = 65 (x + 1)^2$ which clearly has only one solution.