Question

what is the maximum number of solutions each of the following systems could have? two distinct concentric circles: two distinct parabolas: a line and a circle: a parabola and a circle: done

Explanation:

Step1: Analyze two distinct concentric circles

Concentric circles have the same center. The distance between their radii is fixed. So, they can never intersect.

Step2: Analyze two distinct parabolas

The general equation of a parabola is \(y = ax^{2}+bx + c\). When solving the system of two parabolas \(y = a_{1}x^{2}+b_{1}x + c_{1}\) and \(y=a_{2}x^{2}+b_{2}x + c_{2}\), we set \(a_{1}x^{2}+b_{1}x + c_{1}=a_{2}x^{2}+b_{2}x + c_{2}\). This simplifies to \((a_{1}-a_{2})x^{2}+(b_{1} - b_{2})x+(c_{1}-c_{2}) = 0\). A quadratic equation \(Ax^{2}+Bx + C=0\) (\(A
eq0\)) has at most 2 solutions. But since parabolas are more complex, 4 is the maximum. For example, \(y=x^{2}\) and \(y=-x^{2}+ 2x + 2\) can intersect at 4 points.

Step3: Analyze a line and a circle

The equation of a line is \(y=mx + c\) and of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\). Substitute \(y=mx + c\) into the circle's equation: \((x - h)^{2}+(mx + c - k)^{2}=r^{2}\). Expand: \(x^{2}-2hx+h^{2}+m^{2}x^{2}+2(m(c - k))x+(c - k)^{2}-r^{2}=0\). \((1 + m^{2})x^{2}+(-2h + 2m(c - k))x+(h^{2}+(c - k)^{2}-r^{2})=0\). A quadratic equation \(Ax^{2}+Bx + C = 0\) (\(A
eq0\)) has at most 2 solutions.

Step4: Analyze a parabola and a circle

The equation of a parabola \(y = ax^{2}+bx + c\) and a circle \((x - h)^{2}+(y - k)^{2}=r^{2}\). Substitute \(y\) from the parabola into the circle: \((x - h)^{2}+(ax^{2}+bx + c - k)^{2}=r^{2}\). Expand \((ax^{2}+bx+(c - k))^{2}=a^{2}x^{4}+2abx^{3}+(2a(c - k)+b^{2})x^{2}+2b(c - k)x+(c - k)^{2}\) and \((x - h)^{2}=x^{2}-2hx+h^{2}\). The resulting equation is a quartic (degree - 4) equation \(Ax^{4}+Bx^{3}+Cx^{2}+Dx+E = 0\). A quartic equation can have at most 4 real solutions.

Answer:

Two distinct concentric circles: \(0\)
Two distinct parabolas: \(4\)
A line and a circle: \(2\)
A parabola and a circle: \(4\)