praveenvino wrote:
if we join the line connecting the the points (-4,0) and (4,0) to the center of the circle say (0,y), radius will be maximum at the point where the area formed by the above triangle is min. The area will be 0 if the height is 0, which means the center is in the line connecting two pnts (-4,0) and (4,0). Isn't?
The red part is not correct.
Again: The
minimum length of a diameter is indeed 8 (so min r=4) but as ANY point on the y-axis will be equidistant from the given points then any point on it can be the center of the circle thus
the maximum length of the radius is not limited at all.
Check 2 possible circles:
Circle with min radius of 4 (equation x^2+y^2=4^2):
Attachment:
radius 4.gif [ 3.22 KiB | Viewed 74610 times ]
Circle with radius of 5 (equation x^2+(y-3)^2=5^2):
Attachment:
radius 5.gif [ 2.78 KiB | Viewed 74565 times ]
Generally circle passing through the points (4, 0) and (– 4, 0) will have an equation \(x^2+(y-a)^2=4^2+a^2\) and will have a radius of \(r=\sqrt{4^2+a^2}\). As you can see min radius will be for \(a=0\), so \(r_{min}=4\) and max radius is not limited at all (as \(a\) can go to +infinity as well to -infinity).
For more check Coordinate Geometry chapter of Math Book:
math-coordinate-geometry-87652.html Hope it helps.
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