Solve intersection of the straight line with equation $y=x-a$ and the circle equation $x^2 + y^2 = 4$

Question

Find the coordinates of the points of intersection of the straight line with equation $y=x-a$ and the circle with equation $x^2 + y^2 = 4 $. Give the conditions on a for:

I did see the answer from the practice paper: the answer is *see the below

$$\left(\frac{a + \sqrt{8-a^2}}{2}, \frac{-a + \sqrt{8-a^2}}{2} \right); \left(\frac{a - \sqrt{8-a^2}}{2}, \frac{-a - \sqrt{8-a^2}}{2} \right) $$

but I am struggling on how to solve it. can you please explain to me how to solve it. thanks :)

Answer 1

Substituting $y = x - a$ into $x^2 + y^2 = 4$ gives $x^2 + (x - a)^2 = 4$, which expands to $2x^2 - 2ax + a^2 - 4 = 0$. This is a quadratic equation in $x$ that you can then use the quadratic formula to solve to get $x = \frac{2a \pm \sqrt{4a^2 - 4(2)(a^2 - 4)}}{4} = \frac{2a \pm \sqrt{-4a^2 + 32}}{4} = \frac{a \pm \sqrt{8 - a^2}}{2}$, and then get $y$ from $x - a$, giving those $2$ answers you mention.

As for the conditions on $a$, note that you need $8 - a^2 \ge 0$ for $\sqrt{8 - a^2}$ to be a real value, so $a^2 \le 8$ giving $-2\sqrt{2} \le a \le 2\sqrt{2}$ as the range of valid values of $a$.

Answer 2

Here, a figure showing the circle and lines will surely help:

solve

$$x^2+y^2 = 4\ {\rm and} y = x - a$$

so

$$(y+a)^2 + y^2 = 4$$

and use the quadratic equation to get two solutions.