Question

iDentify the equation of the circle that has its center at (-27, 120) and passes through the origin A. (x−27)2+(y+120)2=123 B. (x+27)2+(y−120)2=123 C. (x−27)2+(y+120)2=15129 D. (x+27)2+(y−120)2=15129

Answer 1

The equation of a circle is given in the form $(x-a)^2+(y-b)^2=r^2$, where

(a, b) is the centre of the circle
r = radius

We have the centre of the circle (-27, 120)
We can work out the radius by modelling the radius as the hypotenuse of a triangle as shown in the diagram below

$r^2=27^2+120^2$
$r^2 = 15129$

The equation of the circle is

$[x-(-27)]^2+[y-120]^2=15129$
$[x+27]^2+[y-120]^2=15129$

Correct answer: option number 4