Question 349538
one equation represents a circle and the other equation represents a line.


you can have 0, 1, or 2 solutions.


0 if the line and the circle never intersect
1 if the line is tangent to the circle.
2 if the line intersects the circle at any other angle.


the line can never intersect the circle at more than 2 points.


your equation is:


x^2 + y^2 = 64
x = 1


this is like solving equations simultaneously to get a common solution.


if x = 1, you can substitute in the first equation to get 1 + y^2 = 64 which give you y^2 = 63 which gives you y = +/- sqrt(63).


this implies 2 solutions to this equation.


The graph of the equation of the circle should confirm this.


To graph the equation of the circle, solve for y to get:


y = +/- sqrt(64-x^2)


that graph is shown below:


{{{graph(600,600,-10,10,-10,10,sqrt(64-x^2),-sqrt(64-x^2),7.937253933,-7.937253933))}}}


Draw a vertical line at x = 1 and you will see that the intersection of that vertical line and the circle is around y = +/- 7.937253933.


I drew 2 horizontal lines at y = +/- 7.937253933 so you can see that a little easier.


when x = 0, y = +/- sqrt(64-0) = +/- sqrt(64) = +/- 8.


That's the highest and lowest point of the circle.


when x = 1, y = +/- sqrt(64-1) = +/- sqrt(63) = +/- 7.937253933.  


That's close to the highest point but hot exactly there (just a little below).


Solving the 2 equations simultaneously yielded the result.