Question

1. find the point(s) of intersection of the line y = x + 4 and the circle x² + y² + 6x - 4y = 12.

Explanation:

Step1: Substitute y into circle equation

Substitute $y = x + 4$ into $x^{2}+y^{2}+6x - 4y=12$.
\[

$$\begin{align*} x^{2}+(x + 4)^{2}+6x-4(x + 4)&=12\\ x^{2}+x^{2}+8x + 16+6x-4x-16&=12\\ 2x^{2}+10x-12&=0\\ x^{2}+5x - 6&=0 \end{align*}$$

\]

Step2: Solve the quadratic equation

Factor the quadratic equation $x^{2}+5x - 6 = 0$.
$(x + 6)(x - 1)=0$
So $x=-6$ or $x = 1$.

Step3: Find corresponding y - values

When $x=-6$, $y=-6 + 4=-2$.
When $x = 1$, $y=1 + 4=5$.

Answer:

The points of intersection are $(-6,-2)$ and $(1,5)$.