Recall the equation for a circle with center (

Question

Recall the equation for a circle with center  and radius . At what point in the first quadrant does the line with equation  intersect the circle with radius 5 and center (0, 2)? =  = 

Answer 1

x^2 + (y-2)^2 = 25 is the equation of the circle with radius 5 and center (0,2)

y=2x+2 is the line that intersects the circle twice

x^2 + (2x)^2 = 25 substitute 2x+2 for y, then y-2 = 2x+2-2 = 2x

x^2 +4x^2 = 25

5x^2 = 25

x^2 = 5

x = sqr5 (ignore negative square root, as it's not in the 1st quadrant)

y = 2(sqr5) + 2

point of intersection is (sqr5, 2sqr5 + 2) or about (2.2, 6.4)

Answer 2

x^2 + (y-2)^2 = 25 is the equation of the circle with radius 5 and center (0,2)

y=2x+2 is the line that intersects the circle twice

x^2 + (2x)^2 = 25 substitute 2x+2 for y, then y-2 = 2x+2-2 = 2x

x^2 +4x^2 = 25

5x^2 = 25

x^2 = 5

x = sqr5 (ignore negative square root, as it's not in the 1st quadrant)

y = 2(sqr5) + 2

point of intersection is (sqr5, 2sqr5 + 2) or about (2.2, 6.4)