Final answer:
To find the intersection of the line and the circle, substitute the expression for y from the line's equation into the circle's equation, then solve the resulting quadratic equation. The line y-3x = 10 and the circle x² + y² = 10 intersect at a single point (-3, 1), proving that the line is a tangent to the circle.
Step-by-step explanation:
Intersection of a Line and a Circle
To find the intersection of the line y-3x = 10 and the circle x² + y² = 10, we first solve the line equation for y to get y = 3x + 10. Substituting this into the circle equation gives us x² + (3x + 10)² = 10. Expanding and simplifying, we get a quadratic equation in terms of x, which can be solved to find the point(s) of intersection.
To prove algebraically that these two only intersect once, we need to show that the quadratic equation has only one real solution. After expanding, we have x² + (3x + 10)(3x + 10) - 10 = 0, which simplifies to x² + 9x² + 60x + 100 - 10 = 0, and further to 10x² + 60x + 90 = 0. Factoring out common factors gives us 10(x² + 6x + 9) = 10(x+3)² = 0. This has only one solution for x, which is x = -3. Substituting x = -3 into y = 3(-3) + 10 gives us y = -9 + 10 = 1. The point of intersection is therefore at (-3, 1).
Describing the relationship between the line and the circle with a single word: The line is a tangent to the circle since it touches the circle at exactly one point without crossing it.