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It is possible to draw a circle with diameter x - y - 3 = 0 passing through (2,2) with any radius, but there is a minimum radius that can be drawn when the line through the center of the circle is perpendicular to (2,2). This perpendicular line will have a slope that is the negative reciprocal of the diameter. The slope of the diameter is 1 so the slope of the perpendicular line would be -1/1 = -1.
A line through (2,2) with slope -1 is y - 2 = -(x - 2) or y = -x + 4
y = -x + 4 intercects x - y - 3 = 0 (or y = x - 3) when -x + 4 = x - 3
solving for x we get 2x = 7 or x = 7/2 substituting we get y = 1/2, so the point is at (7/2, 1/2) which you can verify is the only point which is on y = -x + 4 and y = x - 3.
The radius is the distance between the center (7/2, 1/2) and (2,2).
Distance = sqrt((7/2-2)^2 + (1/2 - 2)^2) = sqrt((3/2)^2 + (-3/2)^2) = sqrt(9/4 + 9/4)
Which gives 3/2 sqrt(2) as the minimum radius. Moving the center left or right of this point increases the radius so the minimum radius of a circle with diameter x - y - 3 = 0 and (2,2) a point on the circle is 3/2 sqrt(2).
The center of the circle is the intersection of the line x-y-3 = 0 and parpendicular dropped on it from the point (2,2).
The given line has a slope of +1 by comparing with y = mx +c.
Therefore the parpendicular has a slope given by m1*m2= -1
=> m2 = -1/1= -1
Therefore the parpendicular has equation from slope point formula
(y-y1) = m2(x-x1)
=> y-2 = -1(x-2)
=> x + y -4 =0
The point of intersection of the two lines is found by solving the two equations
x-y-3=0 and x+y-4=0
x=7/2 and y=1/2 are the co-ordinates of the center
Distance of ppoint (2,2) from the center is the radius
=(3/2)sqrt2 .. Ans
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