Circles- tangents and equations

greghunter · #1 $Old$ March 7th, 2009, 03:30 AM

A circle has the equation x^2 + 8x + y^2 + a = 0. A straight line has the equation y= 3^(1/2)x. Find the value of a for which the line is a tangent to the circle.

I thought you might have to use the b^2 - 4ac formula because of the tangent to the circle but I just got stuck.

Hardwarista · #2 $Old$ March 7th, 2009, 04:25 AM

By y= 3^(1/2)x
I suppose you mean $y = 3 - 1 / 2 x$
cause otherwise it wouldn't be a line equation.

Well, you just replace the expression for y into the circle equation, which gives

x^2 + 8x + (3 - 0.5x)^2 + a = 0

After a little bit of manipulation, you will find a 2nd degree polynomial equation in the form

A x^2 + B x + C = 0.

at which point you will - indeed - compute the discriminant D = B^2 - 4 A C, which will be an expression with variable a.
(e.g. : 14 a - 3.5)

Discussion:

if D > 0 there are 2 real solutions for x - i.e. 2 intersection points between the line and the circle
if D<0 there are no real solutions for x - i.e. the line does not 'meet' the circle
if D=0 - 1 real solution - the line is tangent to the circle

So you just solve D=0 to find a.

HallsofIvy · #3 $Old$ March 7th, 2009, 06:09 AM

I would have assumed "y= 3^(1/2)x" mean $y= \sqrt{3}x$ . If that is the case then it is a line through the origin with slope $\sqrt{3}$ . Putting that into the equation of the circle, as Hardwarista suggests, we get $x^2+ 8x+ 3x^2+ a= 0$ or $4x^2+ 8x+ a= 0$ which we can also write as $x^2+ 2x+ a/4= 0$ . If the line is tangent to the circle, this equation must have a single solution so the left side must be a perfect square. Since $(x+1)^2= x^2+ 2x+ 1$ , it's pretty easy to solve for a.