Answer:
The solutions to the system are the points (5, -1) and (1, -5)
Step-by-step explanation:
Hi there!
We have the following system of equations:
x - y = 6
x² + y² = 26
The points common to the line and the circle are those (x,y) values that satisfy both equations. So let´s take the first equation and solve it for x:
x - y = 6
add y to both sides of the equation
x = 6 + y
Now, let´s replace the x in the second equation:
x² + y² = 26
(6 + y)² + y² = 26
(6 + y)(6 + y) + y² = 26
36 + 12y + y² + y² = 26
subtract 26 to both sides of the equation
10 + 12y + 2y² = 0
Solve the quadratic equation using the quadratic formula:
a = 2
b = 12
c = 10
[-b ± √(b² - 4ac)] / 2a
The solutions to the quadratic equation are y = -5 and y = -1
Let´s calculate the x value:
x = 6 + y
For y = -5
x = 6 - 5 = 1
For y = -1
x = 6 - 1 = 5
Then, the solutions to the system and the points at which the line and the circle intersect are (5, -1) and (1, -5). Please, see the attached figure to corroborate this graphically.
Have a nice day!
