Question

which of the following is a degenerate circle?
a. ( x^{2}+y^{2}=1 )
b. ( x^{2}+y^{2}=-1 )
c. ( (x - 4)^{2}+(y - 2)^{2}=0 )
d. ( x + y = 2 )

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.

Step2: Analyze each option

• Option A: \(x^{2}+y^{2}=1\) is a circle with center \((0,0)\) and radius \(r = 1\) (since \(r^{2}=1\)).
• Option B: \(x^{2}+y^{2}=-1\). Since \(x^{2}\geq0\) and \(y^{2}\geq0\) for all real \(x\) and \(y\), \(x^{2}+y^{2}\geq0\). There are no real - valued \(x\) and \(y\) that satisfy this equation. It is not a degenerate circle.
• Option C: \((x - 4)^{2}+(y - 2)^{2}=0\). Since \((x - 4)^{2}\geq0\) and \((y - 2)^{2}\geq0\) for all real \(x\) and \(y\), the only solution is when \(x-4 = 0\) and \(y - 2=0\), i.e., \(x = 4\) and \(y=2\). A degenerate circle (a point - circle) occurs when \(r = 0\) in the standard form \((x - a)^{2}+(y - b)^{2}=r^{2}\). Here \(a = 4\), \(b = 2\) and \(r = 0\).
• Option D: \(x + y=2\) is a linear equation (a straight line) in two variables, not a circle equation.

Answer:

C. \((x - 4)^{2}+(y - 2)^{2}=0\)