Through the point P(3,4) a pair of perpendicular lines are dranw which meet x-axis at the point A and B. The locus of incentre of triangle PAB is

Options

A. `x^(2)-y^(2)-6x-8y+25=0`
B. `x^(2)+y^(2)-6x-8y+25=0`
C. `x^(2)-y^(2)+6x+8y+25=0`
D. `x^(2)+y^(2)+6x+8y+25=0`

Answer

1

Explanation

Let point I be (h,k).
From the figure,
`IP = r cosec 45^(@) , ` where ` r =` inradius
`implies (IP)^(2) =2 r^(2)`
`implies (3-h)^(2)+(4-k)^(2)=2( |k|)^(2)`
( As circle touches x-axis, `r = (|k|)^(2)`
`implies x^(2)-y^(2)-6x-8y+25=0`, which is the require locus.