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The coordinates of the point on the circle x2+y2-12x-4y+30=0 which is the farthest from the origin are
a.(9,3)
b.(8,5)
c.(12,4)
d.none of these
Hi Menka,
The cordiantes, farthest from the origin on the circle, will lie on the line joining the centre of the circle and the point (0,0).
Centre is (6,2).
So eqn of the line is 3y=x.
Sub x=3y in the circle eqn, and solve for y. You will get two values of y (corresponding to which you would get two values of x).
One of the point is the closest point, and the other would be the farthest point.
Best Regards,
Ashwin (IIT Madras).
The ans must be (9,3)
the centre os cirle is (6,2)
and the coordinate which is farthest from line passing through origin is end point of diameter so the line which is passing throung (0,0) has to pass through (6,2) ,i.e it is a secant !
now the equ of line is .
y=1/3x.
not solving the eqn or finding the intersection points of line with circle we put vale of y in the given eqn of circle and get.
x=9,3
y=3,1
obviously 9,3 is farther thn 3,1 .hence 9,3
is ans !
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