Question 207908: Hi! I've been trying to solve this but i have no idea how. Please help me!
find the equation of the circle with center at y-axis and which passes through the origin and the point (4,2)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the equation of the circle with center at y-axis and which passes through the origin and the point (4,2)
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The line thru the origin and (4,2) is a chord of the circle. So the perpendicular bisector of it will go thru the center. The intersection of that line and the y-axis will be the center of the circle.
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Find the eqn of the line thru (0,0) and (4,2).
m = diffy/diffx = 2/4
m = 1/2 (the slope)
The midpoint of the line is (2,1) (see that?)
The slope of the line perpendicular will be the negative inverse, = -2
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Find the eqn of the line with a slope of -2 thru (2,1)
y-y1 = m*(x-x1)
y-1 = -2*(x-2)
y-1 = -2x+4
y = -2x+5
The y-intercept is 5, so the point (0,5) is the center of the circle (it was stated to be on the y-axis).
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Since the circle also goes thru the origin, the radius is 5.
x^2 + (y-5)^2 = 25 is the circle
or
x^2 + y^2 - 10y + 25 = 25
x^2 + y^2 - 10y = 0
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