The center of a circle that includes the points (-5, 2) and (-3, 6) could be (a) (0,2)
The points are given as:
(-5, 2) and (-3, 6)
The equation of a circle is represented as:
[tex](x -a)^2 + (y - b)^2 = r^2[/tex]
Where:
Substitute both points in the above equation
[tex](-5 + a)^2 + (2 - b)^2 = r^2[/tex]
[tex](- 3+ a)^2 + (6 - b)^2 = r^2[/tex]
The radii are equal.
So, we have:
[tex](-5 + a)^2 + (2 - b)^2 = (- 3+ a)^2 + (6 - b)^2[/tex]
Expand the equations
[tex]25 -10a + a^2 + 4 - 4b + b^2 = 9 - 6a + a^2 + 36 - 12b + b^2[/tex]
Evaluate the like terms
[tex]25 -10a + 4 - 4b = 9 - 6a + 36 - 12b[/tex]
Collect like terms
6a -10a - 4b + 12b = 9 + 36 - 25 - 4
Evaluate
-4a + 8b = 16
Divide through by - 4
a - 2b = -4
Next, we test the options:
Option (a) (a,b) = (0,2)
Substitute these values in the equation
0 - 2*2 = -4
Evaluate
-4 = -4
Both sides of the equation are the same
Hence, the center of a circle that includes the points (-5, 2) and (-3, 6) could be (a) (0,2)
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