Find the point(s) of intersection between the circle
x^2 + y^2 = 50 and the linear equation y = 2x − 5.
You could graph the circle and the line and SEE the point(s) of intersection. Or since the line is already re-arranged for y, sub that into the cirle equation and solve for x. And then solve for y : )
thank you John :)
To find the points of intersection between a circle and a linear equation, we need to substitute the equation of the line into the equation of the circle and solve for the values of x and y.
The equation of the circle is x^2 + y^2 = 50.
The equation of the line is y = 2x − 5.
Substituting the equation of the line into the equation of the circle, we get:
x^2 + (2x - 5)^2 = 50.
Expanding the equation, we have:
x^2 + 4x^2 - 20x + 25 = 50.
Combining like terms, we get:
5x^2 - 20x - 25 = 0.
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 5, b = -20, and c = -25. Substituting these values into the formula, we get:
x = (-(-20) ± √((-20)^2 - 4 * 5 * -25)) / (2 * 5).
x = (20 ± √(400 + 500)) / 10.
x = (20 ± √900) / 10.
x = (20 ± 30) / 10.
So we have two possible values for x:
x1 = (20 + 30) / 10 = 5.
x2 = (20 - 30) / 10 = -1.
Now we can substitute these x-values back into the equation of the line to find the corresponding y-values.
For x1 = 5:
y = 2x - 5 = 2 * 5 - 5 = 10 - 5 = 5.
For x2 = -1:
y = 2x - 5 = 2 * (-1) - 5 = -2 - 5 = -7.
Therefore, the points of intersection between the circle and the linear equation are (5, 5) and (-1, -7).