at what point in the first quadrant does the line with equation y=x+2 intersect the circle with radius r=6 and center (-1,0)

Question

at what point in the first quadrant does the line with equation y=x+2 intersect the circle with radius r=6 and center (-1,0) Please show steps

Answer 1

First, we need to write the equation of the circle.  The equation of a circle is

 

(x - h)² + (y - k)² = r²

 

where:

(h, k) is the centerpoint

r = radius

 

So the equation of this circle is then

 

(x + 1)² + y² = 36

 

 

Solving for y gives us

 

y² = 36 - (x + 1)²

 

y = ± √(36 - (x² + 2x + 1))

 

y = ± √(35 - 2x - x²)

 

 

Next, we equate the equation of this circle with the equation of the given line.

 

x + 2 = √(35 - 2x - x²)

 

 

Square both sides of the equation.

 

(x + 2)(x + 2) = 35 - 2x - x²

 

x² + 4x + 4 = 35 - 2x - x²

 

 

Make the right side of the equation equal to zero.

 

2x² + 6x - 31 = 0

 

 

Use the quadratic formula to solve for x:

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 2

b = 6

c = -31

 

 

Plug in these values into the formula.  Remember, since we want the point to be in the first quadrant, choose the positive value of x.  Once you have your positive value of x, substitute it into the equation of the line to get your y value.