(a)
To determine: The possible solutions of a system of one
(a)
Answer to Problem 57E
A quadratic equation and a circle equation can have 0, 1, 2, 3 and 4 solutions.
Explanation of Solution
Calculation:
Consider the quadratic equation be
And the equation of a circle be
Graph 1:
In the above graph, it is seen that a quadratic equation that is, a parabola and a circle equation can intersect at no point.
Graph 2:
In the above graph, it is seen that a quadratic equation that is, a parabola and a circle equation can intersect at one point.
Graph 3:
In the above graph, it is seen that a quadratic equation that is, a parabola and a circle equation can intersect at two points.
Graph 4:
In the above graph, it is seen that a quadratic equation that is, a parabola and a circle equation can intersect at three points.
Graph 5:
In the above graph, it is seen that a quadratic equation that is, a parabola and a circle equation can intersect at four points.
Hence, a quadratic equation and a circle equation can have 0, 1, 2, 3 and 4 solutions.
(b)
To determine: The possible solutions of a system of two equations of circles.
(b)
Answer to Problem 57E
Two equations of circles can have 0, 1, 2, and infinite number of solutions.
Explanation of Solution
Calculation:
Consider the system of two equations of circles be
Graph 1:
In the above graph, it is seen that two equations of circles can intersect at no point.
Graph 2:
In the above graph, it is seen that two equations of circles can intersect at one point.
Graph 3:
In the above graph, it is seen that two equations of circles can intersect at two points.
Graph 4:
In the above graph, it is seen that two equations of circles can intersect at infinite points.
Hence, two equations of circles can have 0, 1, 2, and infinite number of solutions.
Chapter 3 Solutions
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