To find the intersection points of the line and the circle, we first need the equations of both. The line is given by:
y = 2.5x + 4
And the circle with radius 6 and center (0, 4) can be expressed as:
(x – 0)² + (y – 4)² = 6²
which simplifies to:
x² + (y – 4)² = 36
Next, we substitute the line’s equation into the circle’s equation:
x² + (2.5x + 4 – 4)² = 36
which simplifies to:
x² + (2.5x)² = 36
Expanding this gives:
x² + 6.25x² = 36
or
7.25x² = 36
Now, solving for x:
x² = 36 / 7.25
Calculating this, we find:
x² ≈ 4.9655
Taking the square root gives:
x ≈ ±2.224
Since we are looking for the intersection in the first quadrant, we take the positive value:
x ≈ 2.224
To find the corresponding y value, we substitute x back into the line’s equation:
y = 2.5(2.224) + 4
y ≈ 5.56
Thus, the point of intersection in the first quadrant is approximately:
(2.224, 5.56)
In conclusion, the line intersects the circle at the point (2.224, 5.56) in the first quadrant.