Question

question 5 · 1 point
find the intersection of the line and the circle given below.
y = x + 16
x² + y² = 128
give your answer as an ordered pair (x,y). if there are two solutions, give your answer as two ordered pairs separated by a comma. for example, if you found that the solutions were (1,2) and (3,4) you would enter (1,2),(3,4).

Explanation:

Step1: Substitute $y$ into circle equation

Substitute $y=x + 16$ into $x^{2}+y^{2}=128$. We get $x^{2}+(x + 16)^{2}=128$.
Expanding $(x + 16)^{2}$ using $(a + b)^{2}=a^{2}+2ab + b^{2}$, we have $x^{2}+x^{2}+32x+256 = 128$.
Combining like - terms gives $2x^{2}+32x+256 - 128=0$, so $2x^{2}+32x + 128 = 0$.
Dividing through by 2 results in $x^{2}+16x + 64 = 0$.

Step2: Solve quadratic equation

The quadratic equation $x^{2}+16x + 64 = 0$ is in the form $ax^{2}+bx + c = 0$ with $a = 1$, $b = 16$, $c = 64$.
We can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. But this equation is a perfect - square trinomial: $(x + 8)^{2}=0$.
Solving $(x + 8)^{2}=0$ gives $x=-8$.

Step3: Find $y$ - value

Substitute $x=-8$ into $y=x + 16$. Then $y=-8 + 16=8$.

Answer:

$(-8,8)$