Question 192·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

The circle’s center is $(2,3)$, and its radius is $1$ because the right side is $1^2$. Its lowest point is $(2,3-1)=(2,2)$. The parabola’s vertex, its lowest point, is also $(2,2)$ because $(x-2)^2$ is smallest at $x=2$. So both graphs contain $(2,2)$. A point where curves only touch still solves both equations.

Type both equations on separate lines in Desmos. Click the two places where the curves cross: Desmos shows $(1,3)$ and $(3,3)$. The entire circle is in view, so there are no hidden intersections. With the touch at $(2,2)$, the system has exactly three distinct solutions. Choice B.

The first equation gives $y$ in terms of $x$. At a solution, the circle must have that same $y$-value, so replace its $y$ with the entire expression:

Let $u=(x-2)^2$. Since $2-3=-1$, the equation becomes:

This substitution lets you solve for the repeated square before finding $x$.

Multiply $(u-1)$ by itself:

Now it’s an equation you can factor after collecting terms.

Subtract $1$ and combine like terms:

Factor out $u$:

A product is zero when either factor is zero, so $u=0$ or $u=1$.

For $u=0$, take the only possible root:

For $u=1$, take both square roots:

The first equation gives exactly one $y$ for each of these three $x$-values. So the system has exactly three solutions. Choice B.