Problem 81 - Precalculus Essentials

Question

Does \((x-3)^{2}+(y-5)^{2}=-25\) represent the equation of a circle? What sort of set is the graph of this equation?

Step-by-Step Solution

Answer

No, the equation \( (x-3)^{2}+(y-5)^{2}=-25 \) does not represent a circle because the square of the radius is negative. The equation has no real solutions and hence does not represent any real, graphable set.

Step 1: Analyze the General Form

In the equation \((x-3)^{2}+(y-5)^{2}=-25\), it has the same form as the standard equation of a circle, namely \((x-a)^2+(y-b)^{2}=r^2\). However, the = -25 represents a negative radius squared.

Step 2: Reason about the Negative Radius Squared

The square of a distance (in this case the radius) can never be negative because squaring a number always results in a positive value or zero, not negative. Therefore, = -25 is not possible if this was to represent a circle.

Step 3: Determine what the Equation Represents

Given the oddity in the form (a negative square of the radius), the equation \((x-3)^{2}+(y-5)^{2}=-25\) does not represent a circle, or indeed any real graphable shape. Hence, \( (x-3)^{2}+(y-5)^{2}=-25 \) has no real solutions.