The circle passing through(1,-2) and touching the axis of x at (3,0) also passes through ? a)(5,-2) b)(-2,5) c)(-5,2) d)(2,-5)

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Answer 1 · 2018-04-14T06:46:23

#"(a) "(5,-2)# is the Right Choice.

Let, #r gt 0# is the radius of the circle in Question.

Because the circle touches the X-Axis at #(3,0)#, its

centre C must be, #C=C(3,+-r)#.

Given that the point #P=P(1,-2)# lies on the circle, we have,

#PC^2=r^2#.

#:. (1-3)^2+(+-r+2)^2=r^2...[because," Distance Formula]"#.

#:. 4+(r^2+-4r+4)-r^2=0#.

#:. +-4r=-8 rArr r=+-2," but, as "r gt 0, r=+2#.

This means that, the centre is #C=C(3,+-2), and, r=2#.

Hence, the eqns. of the circles are,

# (x-3)^2+(y+-2)^2=2^2,#

#because P(1,-2) cancel(in) (x-3)^2+(y-2)^2=2^2#

#:., "the circle under reference is "(x-3)^2+(y+2)^2=2^2#.

Of all the given points, only #(5,-2)# satisfies the above eqn.

Hence, #"(a) "(5,-2)# is the Right Choice.