Question

Identify the equation of the circle that has its center at (-27, 120) and passes through the origin.

Answer 1

The equation of the circle that has its center at (-27, 120) and passes through the origin is:

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

Solution:

The equation of a circle is given as:

$(x-a)^2+(y-b)^2=r^2$

Where,

(a, b) is the centre of the circle

r is the radius

We have the centre of the circle (-27, 120)

Therefore,

a = -27

b = 120

Given that, it passes through origin. which means, (x, y) = (0, 0)

Substitute (a, b) = (-27, 120) and (x, y) = (0, 0) in eqn

$(0 + 27)^2 + (0 - 120)^2 = r^2\\\\729 + 14400 = r^2\\\\r^2 = 15129$

Substitute $r^2$ = 15129 and (a, b) = (-27, 120) in eqn

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

Thus the equation of circle is found