Question 863824: The centre of a circle lies on the x-axis and passes through the points (4,5) and (-2,3). Find the equation of the circle in the form x^2+y^2+2gx+2fy+c=0
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39623)   (Show Source):
You can put this solution on YOUR website! The center of the circle is some ordered pair, (x,0). The line exactly between the two given points is perpendicular to the line which the two given points defines. The line between the two given points WILL contain the center point of the circle. The x-intercept of this line IS the center of the circle.
The distance formula using the x-intercept and either of the given points will give the radius. Write the circle in standard form, and arrange to the general form you want.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The centre of a circle lies on the x-axis and passes through the points (4,5) and (-2,3). Find the equation of the circle in the form x^2+y^2+2gx+2fy+c=0
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Plot the points; let the center be (x,0)
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Radii are equal::
sqrt[(4-x)^2 + (5-0)^2] = sqrt[(-2-x)^2+(3-0)^2]
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(4-x)^2 + 5^2 = (2+x)^2+9
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16-8x+x^2 + 25 = 4 + 4x + x^2 + 9
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41 - 6x = 13+4x
10x = 28
x = 2.8
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Center is (2.8,0)
radius = sqrt[(4-2.8)^2 + 25] = sqrt[1.44+ 25] = sqrt(26.44)
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Equation:
(x-2.8)^2 + y^2 = 26.44
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Cheers,
Stan H.
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