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6 days ago

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

Mathematics

1 answer:

zzz [4K]6 days ago

6 0

The equation representing the circle centered at (-27, 120) that passes through the origin is:

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

Solution:

The general equation of a circle is expressed as:

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

Where,

(a, b) denotes the center of the circle

r signifies the radius

Given the center as (-27, 120)

Thus;

a = -27

b = 120

Considering it intersects the origin, meaning (x, y) = (0, 0)

Substituting (a, b) = (-27, 120) and (x, y) = (0, 0) into the equation

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

Input $r^2$ = 15129 and (a, b) = (-27, 120) into the equation

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

Hence, the equation characterizing the circle is determined

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