1. ## intersecting circles

Hi folks,

I am trying to find the intersection points A and B of the two circles shown in the attachment. I am getting imaginary roots, which I was not expecting.

The equations of the circles are:

$x^2 + y^2 - 4x = 0$ and

$x^2 + y^2 -4y + 3 = 0$

subtracting one from the other gives $-4x + 4y - 3 = 0 \Rightarrow y = x + \frac{3}{4}$

substituting in the first circle: $x^2 + (x + \frac{3}{4})^2 - 4x = 0$

$2x^2 + \frac{3}{2}x + \frac{9}{4} - 4x = 0 \Rightarrow 2x^2 - \frac{5}{2}x + \frac{9}{4} = 0$

$8x^2 - 10x + 9 = 0$

a quadratic in x with a = 8, b = -10 and c = 9 produces a discriminant $(b^2 - 4ac) < 0$ i.e. no real roots.

Clearly I am making a mistake somewhere! Can anyone see it?

3. ## Re: intersecting circles

$\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16} \neq \dfrac{9}{4}$

many thanks.

5. ## Re: intersecting circles

Originally Posted by SlipEternal
$\left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16} \neq \dfrac{9}{4}$
Sir Ingram, go stand in the corner for 9/16 hour.....

6. ## Re: intersecting circles

You're right! I am deeply ashamed.