If (5, 4) and (7,2) are the endpoints of a diameter of a circle, what is the equation of the circle?
2 answers:
Answer:
(x -6)^2 +(y -3)^2 = 2
Step-by-step explanation:
The midpoint (C) is the center of the circle:
C = ((5, 4) +(7, 2))/2 = ((5+7)/2, (4+2)/2) = (6, 3)
A circle with center (h, k) that goes through point (a, b) can be written as ...
(x -h)^2 +(y -k) ^2 = (a -h)^2 +(b -k)^2
Using the values for this problem, we get ...
(x -6)^2 +(y -3)^2 = (5 -6)^2 +(4 -3)^2
(x -6)^2 +(y -3)^2 = 2
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<em>Example: (View attached image below). </em><em>Table A</em><em> is a function because each </em><em>x </em><em>value has only 1 </em><em>y</em><em> value. But </em><em>Table B</em><em> is not a function because the </em><em>x value</em><em> of </em><em>4 </em><em>has </em><em>2 y values</em><em>.</em>