Answer:
C: there are two real solutions
Step-by-step explanation:
x^2 + y^2 = 36 is the equation of a circle centered at the origin and with radius √36, or 6. Even a rough sketch of this circle would enable you to visualize what is happening here.
3x – y + 1 = 0 may be solved for y: y = 3x + 1. The y-intercept of this straight line is (0, 1). Plot this point inside the circle of radius 6 mentioned above and then draw a straight line with slope m = 3 through it. This line will intersect the circle in two places; they represent real solutions.
Alternatively, substitute 3x + 1 for y in x^2 + y^2 = 36:
x^2 + (3x + 1)^2 = 36, or
x^2 + 9x^2 + 6x + 1 = 36, or
10 x^2 + 6x + 1 - 36 = 0, or
10 x^2 + 6x - 35 = 0. This is a quadratic equation. We'll use the quadratic formula to find solutions which represent the intersections of this line with this circle:
The coefficients are a = 10, b = 6 and c = -35. Thus, the discriminant is
b^2 - 4ac, or 6^2 - 4(10)(-35), or 1436.
Because the discriminant is positive, we can safely conclude that there are two real solutions, that is, two different, real x-values, each representing the x-coordinate of a point of intersection of the circle and the line.
The correct answer to this prolem is C: there are two real solutions.
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