Answer:
The point of intersection would be (3,-2)
Step-by-step explanation:
To determine the intersection of the conics we can use a system of equations since they intersect when they are both equal.
Then, we have these equations:
[tex]\begin{gathered} (x+1)^2+(y+2)^2=16\text{ (1)} \\ (y+2)^2=16-(x+1)^2\text{ (1)} \\ (y+2)^2=4(x-3)\text{ (2)} \end{gathered}[/tex]
Equalize equations (1) and (2).
[tex]\begin{gathered} 4(x-3)=16-(x+1)^2 \\ \end{gathered}[/tex]
Solve for the x-coordinate.
[tex]\begin{gathered} 4x-12=16-(x^2+2x+1) \\ 4x-12=16-x^2-2x-1 \\ x^2+6x-27=0 \\ (x-3)(x+9)=0 \\ \text{Possible x-intersections:} \\ x=3 \\ x=-9 \end{gathered}[/tex]
Since the circle has a radius of 4, we know that the intersection cannot be x=-9. Then, the x-coordinate to use is x=3.
Substitute x=3 into one of the equations to determine the y-coordinate:
[tex]\begin{gathered} (y+2)^2=4(3-3) \\ (y+2)^2=0 \\ y^2+4y+4=0 \\ (y+2)(y+2)=0 \\ y-\text{coordinate:} \\ y=-2 \end{gathered}[/tex]
Hence, the point of intersection would be (3,-2)
