We are given the following system of equations:
[tex]\begin{gathered} (x+1)^2+(y+2)^2=16,\text{ (1)} \\ (y+2)^2=4(x-3),\text{ (2)} \end{gathered}[/tex]We are asked to find the point of interception or equivalently find the solution to the system. To do that, we will replace the value of (y+2) from equation (2) into equation (1), like this:
[tex](x+1)^2+4(x-3)=16[/tex]Now we will solve the parenthesis. For the parenthesis on the left we'll use the following property:
[tex](a+b)^2=a^2+2ab+b^2^{}_{}[/tex]Solving we get:
[tex]x^2+2x+1+4x-12=16[/tex]Adding like terms:
[tex]x^2+6x-11=16[/tex]Now we will subtract 16 on both sides:
[tex]\begin{gathered} x^2+6x-11-16=0 \\ x^2+6x-27=0 \end{gathered}[/tex]Now we will solve this equation using the quadratic formula, that is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where:
[tex]\begin{gathered} a=1 \\ b=6 \\ c=-27 \end{gathered}[/tex]Replacing we get:
[tex]x=\frac{-6\pm\sqrt[]{6^2-4(1)(-27)}}{2(1)}[/tex]Now we solve the operations:
[tex]x=\frac{-6\pm\sqrt[]{144}}{2}[/tex][tex]x=\frac{-6\pm12}{2}[/tex][tex]x_1=\frac{-6+12}{2}=3[/tex][tex]x_2=\frac{-6-12}{2}=-9[/tex]Now we replace the values of "x" in equation (2)
[tex](y+2)^2=4(x-3)[/tex]For x = 3
[tex]\begin{gathered} (y+2)^2=4(3-3) \\ (y+2)^2=0 \end{gathered}[/tex]Now we solve for "y", by taking square roots on both sides:
[tex]\begin{gathered} (y+2)=0 \\ y=-2 \end{gathered}[/tex]For x = -9
[tex]\begin{gathered} (y+2)^2=4(-9-3) \\ (y+2)^2=4(-12) \end{gathered}[/tex]Taking square root:
[tex]\begin{gathered} (y+2)=\sqrt[]{4(-12)} \\ (y+2)=\sqrt[]{-48\text{ }} \end{gathered}[/tex]Since we get the square root of a negative number there are no real "y" solution for x = -9, therefore, the only real solution is (x,y)=(3,-2)
