Question

What's the center of a circle whose equation is (x – 14)2 + (y + 21)2 = 64?

Question 19 options:


1)
(–14, –21)


2)
(14, –21)


3)
(–14, 21)


4)
(14, 21)

Answer 1

The center of the circle is:

• 
  `\left(a,\:b\right)=\left(14,\:-21\right)`

Thus, option (2) is true.

Explanation:

The circle equation is given by

`\left(x-a\right)^2+\left(y-b\right)^2=r^2`

here,

• r = raduis

• center = (a, b)

Given the equation

`\left(x-14\right)^2+\left(y+21\right)^2=64`

`\mathrm{Rewrite}\:\left(x-14\right)^2+\left(y+21\right)^2=64\:\mathrm{in\:the\:form\:of\:the\:standard\:circle\:equation}`

`\left(x-14\right)^2+\left(y-\left(-21\right)\right)^2=8^2`

comparing with the circle equation

`\left(x-a\right)^2+\left(y-b\right)^2=r^2`

Therefore, the center of the circle is:

• 
  `\left(a,\:b\right)=\left(14,\:-21\right)`

Thus, option (2) is true.