the equation x^2+y^=25 describes a circle with center at the origin and radius 5. The line y=x-1 passes through the circle. Using the substitution method, find the points at which the circle and line intersect.
a) (4,3) and (-4,-3)
b) (3,4) and (-3,-4)
c) (4,-3) and (-3,4)
d) no solution
well, do the sub:
x^2 + (x-1)^2 = 25
or, even without doing the math, you can see that none of the choices satisfies y=x-1 for both points
But (d) is also wrong, since the line crosses at (-3,-4) and (4,3)
I suspect at least one typo.
Thanks.
That was my problem, I couldn't find a right answer.
Is there a problem with my work?
x^2+y^2=25 y=x-1
x^2+(x-1)^2=25 (x-1)(x-1)=x^2-2x+1
x^2+x^2-2x+1=25
2x^2-2x+1=25
2x^2-2x-24
2(x^2-x-12)
2(x-4)(x+3)
x-4) (x+3)
x=4 x=-3
y=x-1
y=4-1 y=-3-1
y=3 y=-4
(4,3) (-3,-4)
Looks good to me
To find the points at which the circle and the line intersect, we can solve the system of equations formed by the equation of the circle, x^2 + y^2 = 25, and the equation of the line, y = x - 1.
Start by substituting y in the equation of the circle with the expression for y in terms of x from the equation of the line:
x^2 + (x - 1)^2 = 25
Expanding the equation:
x^2 + (x^2 - 2x + 1) = 25
Combining like terms:
2x^2 - 2x - 24 = 0
Dividing both sides by 2:
x^2 - x - 12 = 0
This equation is a quadratic equation that can be factored. Factoring, we have:
(x + 3)(x - 4) = 0
Setting each factor equal to zero gives us two possible values for x:
x + 3 = 0 --> x = -3
x - 4 = 0 --> x = 4
Now, substitute these values of x back into the equation of the line to find the corresponding values of y:
For x = -3:
y = -3 - 1 --> y = -4
For x = 4:
y = 4 - 1 --> y = 3
Therefore, the points of intersection are (-3, -4) and (4, 3), which correspond to option a) (4,3) and (-4,-3).