Question

There is one point of intersection in the graphs of the conics

Answer 1

We are given the following system of equations:

`\begin{gathered} (x+1)^2+(y+2)^2=16,\text{ (1)} \\ (y+2)^2=4(x-3),\text{ (2)} \end{gathered}`

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:

`(x+1)^2+4(x-3)=16`

Now we will solve the parenthesis. For the parenthesis on the left we'll use the following property:

`(a+b)^2=a^2+2ab+b^2^{}_{}`

Solving we get:

`x^2+2x+1+4x-12=16`

Adding like terms:

`x^2+6x-11=16`

Now we will subtract 16 on both sides:

`\begin{gathered} x^2+6x-11-16=0 \\ x^2+6x-27=0 \end{gathered}`

Now we will solve this equation using the quadratic formula, that is:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where:

`\begin{gathered} a=1 \\ b=6 \\ c=-27 \end{gathered}`

Replacing we get:

`x=\frac{-6\pm\sqrt[]{6^2-4(1)(-27)}}{2(1)}`

Now we solve the operations:

`x=\frac{-6\pm\sqrt[]{144}}{2}`
`x=(-6\pm12)/(2)`
`x_1=(-6+12)/(2)=3`
`x_2=(-6-12)/(2)=-9`

Now we replace the values of "x" in equation (2)

`(y+2)^2=4(x-3)`

For x = 3

`\begin{gathered} (y+2)^2=4(3-3) \\ (y+2)^2=0 \end{gathered}`

Now we solve for "y", by taking square roots on both sides:

`\begin{gathered} (y+2)=0 \\ y=-2 \end{gathered}`

For x = -9

`\begin{gathered} (y+2)^2=4(-9-3) \\ (y+2)^2=4(-12) \end{gathered}`

Taking square root:

`\begin{gathered} (y+2)=\sqrt[]{4(-12)} \\ (y+2)=\sqrt[]{-48\text{ }} \end{gathered}`

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)