Algebraic Approach: Solving the Equations

How do we solve two equations in two unknowns? We eliminate one of the variables, giving us one equation in one unknown and solve that. So examine the two equations. Which variable should we eliminate? Well it doesn’t matter much. Either way we get the same kind of equation:

Eliminate y: x2 + (x2 − t)2 = 1
Eliminate x: y + (y − t)2 = 1

Let’s go with the second equation. It’s a lower degree. This is a quadratic equation in y which means we can put it in standard form:

y + (y2 - 2yt + t2) = 1
y2 - (2t - 1)y + (t2 - 1) = 0
And then we can use the quadratic formula to find the roots:

 
 

Okay. How many solutions do we have? The answer depends on the sign of the discriminant D = 5 - 4t.

• D < 0 no solutions, t > 5/4
• D = 0 one solution , t = 5/4
• D > 0 two solutions, t < 5/4

Our critical value t* is the case of one solution (two intersections with the same y-value). D = 5 - 4t = 0 ⇒ t∗ = 5/4

This is the centre of the circle of tangency, and from the diagram that looks about right. The two values of x are symmetrically placed about the y-axis. Indeed since y = x2:

 
 

and the two x-values have opposite sign.
To find the intersection points put t = 5/4 into the formula for y:

 

and then the x-values are:

Which means the final intersections are (1.73/2, 3/4) and (-1.73/2, 3/4)