The circle passing through (1, -2) and touching the axis of x at (3, 0) also passes through the point (1) (2, -5) (2) (5, -2) (3) (-2, 5) (4) (-5, 2)

Options

A. (-5, 2)
B. (2, -5)
C. (5, -2)
D. (-2, 5)

Answer

C

Explanation

Let the equation of circle be
`(x-3)^(2)+(y-0)^(@)+lambday=0`
` (##41Y_SP_MATH_C16_E04_005_S01.png" width="80%">
As it passes through (1, -2)
`therefore" " (1-3)^(2)+(-2)^(2)+lambda(-2)=0`
`rArr " " 4+4-2lambda=0rArrlambda=4`
`therefore` Equation of circle is
`(x-3)^(2)+y^(2)+4y=0`
by hit and trial method, we are that point (5, -2) satisfies equation of circle.