1. ## tangent

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

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

thanks

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

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

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

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

thanks
Thankyou for showing all your work. It makes it very simple to help you.

The simplification should be $65x^2 + 130x + 65 = 0$.

This in turn is equivalent to $0 = 65 (x^2 + 2x + 1) = 65 (x + 1)^2$ which clearly has only one solution.

3. ah of course, stupid mistake there. thanks