Question

16 mark for review which of the following equations represents a circle in the xy - plane that intersects the y - axis at exactly one point? (a) (x - 8)^2+(y - 8)^2 = 16 (b) (x - 8)^2+(y - 4)^2 = 16 (c) (x - 4)^2+(y - 9)^2 = 16 (d) x^2+(y - 9)^2 = 16

Explanation:

Step1: Recall circle - equation and y - axis intersection

The standard form of a circle equation is \((x - a)^2+(y - b)^2=r^2\), where \((a,b)\) is the center of the circle and \(r\) is the radius. To find the intersection with the \(y\) - axis, set \(x = 0\).

Step2: Analyze option A

Set \(x = 0\) in \((x - 8)^2+(y - 8)^2=16\), we get \((0 - 8)^2+(y - 8)^2=16\), i.e., \(64+(y - 8)^2=16\), \((y - 8)^2=- 48\), no real - solutions.

Step3: Analyze option B

Set \(x = 0\) in \((x - 8)^2+(y - 4)^2=16\), we get \((0 - 8)^2+(y - 4)^2=16\), \(64+(y - 4)^2=16\), \((y - 4)^2=-48\), no real - solutions.

Step4: Analyze option C

Set \(x = 0\) in \((x - 4)^2+(y - 9)^2=16\), we get \((0 - 4)^2+(y - 9)^2=16\), \(16+(y - 9)^2=16\), \((y - 9)^2=0\), \(y = 9\), one solution.

Step5: Analyze option D

Set \(x = 0\) in \(x^2+(y - 0)^2=16\), we get \(0+(y - 0)^2=16\), \(y^2=16\), \(y=\pm4\), two solutions.

Answer:

C. \((x - 4)^2+(y - 9)^2=16\)