Question

Recall the equation for a circle with center (h,k) and radius r. At what point in the first quadrant does the line with equation y=x+4 intersect the circle with radius 3 and center (0, 4)?

Answer 1

Explanation:

the circle equation is

r² = (x - h)² + (y - k)²

in our case

3² = (x - 0)² + (y - 4)²

9 = x² + y² - 8y + 16

x² + y² - 8y + 7 = 0

this intersects with

y = x + 4

that means we have

x² + (x + 4)² - 8(x + 4) + 7 = 0

x² + x² + 8x + 16 - 8x - 32 + 7 = 0

2x² - 9 = 0

2x² = 9

x² = 9/2

x = 3/sqrt(2) = 2.121320344...

y = 3/sqrt(2) + 4 = 6.121320344...

the point is

(3/sqrt(2), 3/sqrt(2) + 4) = (2.121320344..., 6.121320344...)