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 · 2016-11-30T14:39:32

Let the equation be $x^{2} + y^{2} + A x + B y + C = 0$ $x^2 + y^2 + Ax + By + C = 0$

Accordingly, we can write a system of equations.

Equation #1:

${(- 9)}^{2} + {(- 16)}^{2} + A (- 9) + B (- 16) + C = 0$ $(-9)^2 + (-16)^2 + A(-9) + B(-16) + C = 0$

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

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

Equation #2

${(- 9)}^{2} + {(32)}^{2} - 9 A + 32 B + C = 0$ $(-9)^2 + (32)^2 - 9A + 32B + C = 0$

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

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

Equation #3

${(22)}^{2} + {(15)}^{2} + 22 a + 15 B + C = 0$ $(22)^2 + (15)^2 + 22a + 15B + C = 0$

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

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!