The radius of the circle which has normals `xy-2x-y+2= 0` and a tangent `3x+4y-6= 0` is

Options

A. `x^(2)+y^(2)-2x-4y+4=0`
B. `x^(2)+y^(2)-2x-4y+5=0`
C. `x^(2)+y^(2)=5`
D. `(x-3)^(2)+(y-4)^(2)=5`

Answer

1

Explanation

The two normals are `x=1` and `y=2`.
Their point of intersection (1,2) is the center of the required circle.
Radius ` (|3+8-6|)/(5)=1`
Therefore, the required circle is
`(x-1)^(2)+(y-2)^(2)=1`
i.e., `x^(2)+y^(2)-2x-4y+4=0`