The equation of a circle which has a tangent `3x+4y=6` and two normals given by `(x-1)(y-2)=0` is

Options

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

Answer

C

Explanation

`(x-1)(t-2)=0`
`impliesx-1=0 and y-2=0`
`therefore` Radius `=(3(1)+4(2)-6)/(sqrt(9+16))`
`=5/5=1`
`therefore` Equation of the circle is
`(x-1)^(2)+(y-2)=1`
or `x^(2)+y^(2)-2x-4y+4=0`