Question

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

Answer 1

Answer:

The center of the circle is:

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

Thus, option (2) is true.

Step-by-step 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.