Question

Identify the equation of the circle that has its center at (-16, 30) and passes through the origin.A. (x−16)^2+(y+30)^2=34B. (x−16)^2+(y+30)^2=1156C. (x+16)^2+(y−30)^2=1156D. (x+16)^2+(y−30)^2=34

Answer 1

In order to determine the equation of the given circle, take into account the general formula for the equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

where the point (h,k) is the center of the circle.

You have tha the center of the circle is (-16,30). It means that:

h = -16

k = 30

then, the left side of the equation becomes:

`\begin{gathered} (x-(-16))^2+(y-30)^2 \\ =(x+16)^2+(y-30)^2 \end{gathered}`

next, consider that if the point passestroug the origin, it is necessary that for

x = 0 and y = 0 the equation coincides with one of the given options.

Replace x=0 and y=0 in the following expression:

`(0+16)^2+(0-30)^2=1156`

Hence, it is necessary taht r^2 = 1156

The requierd equation of the circle is:

`(x+16)^2+(y-30)^2=1156`