Math 127 help please

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+2 intersect the circle with radius 3 and center (0, 2)? 

x = Incorrect 

y = 

Thank your for you help and time

Answer 1

(x-0)² + (y-2)² = 9

y = x+2

x² + (x+2-2)² = 9

2x² = 9

x² = 9/2

So, x = 3√2/2 and y = x+2 = (3√2+4)/2

Answer 2

radius 3, center (0,2) is the circle with equation

x^2 + (y-2)^2 = 3^2 = 9

the line y =x+2 intersects the circle when

x^2 +(x+2-2)^2 = 9

2x^2 = 9

x^2 = 9/2

x = +/- 3/sqr2 = +/-1.5sqr2 and y =+/-1.5sqr2 +2

the points

(1.5sqr2, 1.5sqr2 +2) and (-1.5sqr2, -1.5sqr2 +2)

one intersection point in quadrant I, the other in quadrant III

= about (2.1, 4.1) and (-2.1, -0.1)

really helps to sketch a graph of the circle and line

although you can do it all algebraically

you can see what's going on better with a graph

Answer 3

The equation being recalled is as follow:

(x² - h) + (y² - k) = r

If we are only interested with values in the first quadrant, then we will only accept solutions where x > 0.

Given the specifics of the circle, we want to find where the lines x² + (y² - 2) = 3 and y = x + 2 intersect.

The best way to determine where these lines intersect is to substitute one equation into the other. I'd recommend substituting what y is equal to, in the latter equation, into the former equation, then solve for x. After that you can plug the solution for x into either equation to get the corresponding y for the coordinate at which they intersect.