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: Use the distance formula concept

If a point $(a,b)$ lies on the circle, then $(a + 4)^2+(b - 19)^2 = 121$. We can also consider the range of $x$ - values. The $x$ - values of points on the circle satisfy $|x-(-4)|\leq11$, i.e., $-4 - 11\leq x\leq-4 + 11$, or $-15\leq x\leq7$.

Step3: Check each option

• Option A: If $a=-16$, $| - 16+4|=| - 12| = 12>11$, so the point is outside the circle.
• Option B: If $a=-14$, $| - 14 + 4|=| - 10|=10\leq11$, so the point can lie on the circle.
• Option C: If $a = 11$, $|11 + 4|=15>11$, so the point is outside the circle.
• Option D: If $a = 19$, $|19+4|=23>11$, so the point is outside the circle.

Answer:

B. - 14