If the line `4x - 3y = -12` is tangent at point `(-3,0)` and the line `3x+4y=16` is tangent at the point `(4,1)` to a circle then equation of circle

Options

A. `x^(2)+y^(2)-2x+6y-15=0`
B. `x^(2)+y^(2)-2x+6y-20=0`
C. `x^(2)+y^(2)+2x+6y-15=0`
D. `x^(2)+y^(2)-2x-6y-15=0`

Answer

A

Explanation

Clearly, centre C of the required circle is the point of intersection of the normals at A and B.
The equation of a line through A (-3, 0) and perpendicular to 4x-3y=-12 is
`y-0=(-3)/(4)(x+3)rArr 3x+4y+9=0 " " ...(i)`

Similarly, the equation of a line through B(4, 1) and perpendicular to 3x+4y=16 is
`y-1=(4)/(3)(x-4)rArr 4x-3y-13=0 " " ...(ii)`
Solving (i) and (ii), we get x=1, y=-3.
So, the coordinates of C are (1, -3).
`:.` Radius ` = CA=sqrt(16+9)=5`.
Hence, the equation of the required circle is
`(x-1)^(2)+(y+3)^(2)=5^(2)` or, `x^(2)+y^(2)-2x+6y-15=0`