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)

1 Answer
Apr 14, 2018

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

Explanation:

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.