Matthew C. answered • 08/19/21
10+ years Experienced Math Tutor, Georgia Tech Grad and Data Scientist
Remember in Algebra solving where two lines intersect? You found two equations and were able to solve for the (x,y) point that they intersect by solving the system of equations for x and y. We're going to do the same thing here. We have the equation of a line, now let's get the equation of a circle.
The general equation of a circle is (x-h)2 + (y-k)2 = r2. h and k are the x and y coordinates of the center of the circle, respectively.
With (h,k) = (0,3), and r = 6, we have:
(x - 0)2 + (y-3)2 = 62
After simplifying we get:
x2 + (y-3)2 = 36
Now we need to do some substitution. We know that y = 2x + 3, so let's substitute (2x + 3) in for y in the circle equation:
x2 + (2x + 3 - 3)2 = 36
Simplifying:
x2 + (2x)2 = 36
5x2 = 36
We can solve for x by dividing by 5 then taking the square root of both sides.
x = +/- √(36/5) = +/- 6/√(5)
We generally don't like having square roots in the denominator. So we're going to "rationalize" the denominator by multiplying the numerator and the denominator by √(5).
x = +/- 6√(5)/5
We have two answers for x because of the plus or minus. The question specifies that we're looking for a point of intersection in the first quadrant. Points in the first quadrant have a positive x and a positive y. Therefore, x = + 6√(5)/5
Now we can substitute x to solve for y (why not use the easier equation?):
y = 2x + 3
y = 2(6√(5)/5) = 12√(5)/5
Therefore, the point of intersection between the line and the circle in the first quadrant is:
(6√(5)/5, 12√(5)/5)
