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A circle is centered at the point (5, -4) and passes through the point (-3, 2).
The equation of this circle is (x +_ )2 + (y +_ )2 = _

Answer :

Answer:

The answer to your question is    (x - 5)² + (x + 4) = 100

Step-by-step explanation:

Data

Center = (5, -4)

Point = (-3, 2)

Process

1.- Calculate the length of the radius

Formula

      d = [tex]\sqrt{(x2-x1)^{2} + (y2 - y1)^{2}}[/tex]

Substitution

     d = [tex]\sqrt{(-3-5)^{2}+ (2 + 4)^{2}}[/tex]

Simplification

     d = [tex]\sqrt{(-8)^{2}+ (6)^{2}}[/tex]

     d= [tex]\sqrt{64 + 36}[/tex]

     d = [tex]\sqrt{100}[/tex]

     d = 10

2.- Get the equation of the line

     h = 5    k = -4    r = 10

     ( x - 5)² + (y + 4)² = 10²

Simplification and result

                               (x - 5)² + (x + 4) = 100

andrewinsoul

Answer:

The first space is 3,

The second space is -2

The third space is 100

Step-by-step explanation:

A circle with center (a, b) and has radius r has equation: [tex](x-a)^{2} + (y-b)^{2} = r^{2}[/tex]

Now, if the circle passes through (-3, 2) and it has center on (5, -4). That means the radius of the circle will be the distance between points (-3, 2) and (5, -4).

[tex]d = \sqrt{(x_{2} - x_{1} )^{2} + (y_{2} - y_{1} )^{2} }[/tex]

where d = formula of distance between points [tex](x_{2}, y_{2} ) \ and \ (x_{1}, y_{1} )[/tex]

[tex]d = \sqrt{(-3-5)^{2} + (-4-2)^{2} }\\ d = \sqrt{-8^{2} + -6^{2} }\\ d = \sqrt{64+36}\\ d = \sqrt{100}\\ d = radius \ of \ circle = 10[/tex]

Now, with r = 10, a = -3, b = 2

The equation of circle becomes:

[tex](x-(-3))^{2} + (y-2)^{2} = 10^{2} \\(x+3)^{2} + (y-2)^{2} = 100 \\[/tex]

The first space is 3,

The second space is -2

The third space is 100

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