Why is clinical thermometer not suitable to measure temperature of boiling water?
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Slope = (y2 - y1)/(x2 - x1) = (1 - 3)/(-2 - (-5)) = -2/(-2 + 5) = -2/3
For perpendicular lines, m2 = -1/m1 = -1/(3/8) = -8/3
A line passing through point (-4, 0) and parallel to the y-axis is x = -4
radius = diameter/2 = 10/2 = 5
center of circle = midpoint of diameter = ((-3 + 5)/2, (1 + 7)/2) = (2/2, 8/2) = (1, 4)
Required equation is (x - 1)^2 + (y - 4)^2 = 5^2
(x - 1)^2 + (y - 4)^2 = 25
For perpendicular lines, m2 = -1/m1 = -1/(3/8) = -8/3
A line passing through point (-4, 0) and parallel to the y-axis is x = -4
radius = diameter/2 = 10/2 = 5
center of circle = midpoint of diameter = ((-3 + 5)/2, (1 + 7)/2) = (2/2, 8/2) = (1, 4)
Required equation is (x - 1)^2 + (y - 4)^2 = 5^2
(x - 1)^2 + (y - 4)^2 = 25