What is the standard form of the equation of a circle passes through the points (–9, –16), (–9, 32), and (22, 15)?

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Answer 1

Let the equation be #x^2 + y^2 + Ax + By + C = 0#

Accordingly, we can write a system of equations.

Equation #1:

#(-9)^2 + (-16)^2 + A(-9) + B(-16) + C = 0#

#81 + 256 - 9A - 16B + C = 0#

#337 - 9A - 16B + C = 0#

Equation #2

#(-9)^2 + (32)^2 - 9A + 32B + C = 0#

#81 + 1024 - 9A + 32B + C = 0#

#1105 - 9A + 32B + C = 0#

Equation #3

#(22)^2 + (15)^2 + 22a + 15B + C = 0#

#709 + 22A + 15A + C = 0#

The system therefore is #{(337 - 9A - 16B + C = 0),(1105 - 9A + 32B + C = 0), (709 + 22A + 15B + C = 0):}#

After solving, either using algebra, a C.A.S (computer algebra system) or matrices, you should get solutions of #A = 4, B = -16, C = -557#.

Hence, the equation of the circle is #x^2 + y^2 + 4x - 16y -557 = 0#.

Hopefully this helps!