Question

20 mark for review
(x + 4)^2+(y - 19)^2 = 121
the graph of the given equation is a circle in the xy - plane. the point (a,b) lies on the circle. which of the following is a possible value for a?
a -16
b -14
c 11
d 19

Explanation:

Step1: Recall circle - equation form

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. For the equation \((x + 4)^2+(y - 19)^2 = 121\), the center is \((-4,19)\) and the radius \(r=\sqrt{121}=11\).

Step2: Substitute \(x = a\) into the circle equation

Since the point \((a,b)\) lies on the circle, we have \((a + 4)^2+(b - 19)^2 = 121\). Then \((a + 4)^2=121-(b - 19)^2\). Since \((b - 19)^2\geq0\), we know that \((a + 4)^2\leq121\), so \(- 11\leq a + 4\leq11\).

Step3: Solve the inequality for \(a\)

Subtract 4 from all parts of the inequality: \(-11-4\leq a+4 - 4\leq11 - 4\), which simplifies to \(-15\leq a\leq7\).

Answer:

B. - 14