lf a circle `C` passing through `(4,0)` touches the circle `x^2 + y^2 + 4x-6y-12 = 0` externally at a point `(1, -1),` then the radius of the circle `C` is :-

Options

A. 5
B. `2sqrt5`
C. `sqrt57`
D. 4

Answer

A

Explanation

Equation of tangent to the circle
`x^(2)+y^(2)+4x-6y-12=0 " at " (1,-1)` is given by
`"xx"_(1)+yy_(1)+2(x+x_(1))-3(y+y_(1))-12=0, " where " x_(1)= 1 and y_(1) = -1`
`rArrx-y+2(x+1)-3(y-1)-12=0`
`rArr 3x - 4y-7=0`
This will also a tangent to the required circle.
Now, equation of family of circles touching the line
`3x-4y-7=0 " at point " (1,-1)` i s given bt
`
(x-1)^(2)+(y+1)^(2)+lambda (3x-4y-7)=0`
So, the equation of required circle will be
`(x-1)^(2)+(y+1)^(2)+lambda(3x-4y-7)=0, " for some " lambdainR" "...(i)`
`therefore` The required circle passes through (4, 0)
`therefore(4-1)^(2)+(0+1)^(2)+lambda(3xx4-4xx0-7)=0`
`rArr 9+1+lambda(5)=0rArrlambda=-2`
Subtituting `lambda = -2` in Eq. (i), we get
`(x-1)^(2)+(y+1)^(2)-2(3x-4y-7)=0`
`rArr x^(2) + y^(2)-8x+10y+16=0`
On comparing it with
`x^(2)+y^(2) +2gx+2fy+c=0`, we get
`g=-4, f=5,c=16`
`therefore "Radius"=sqrt(g^(2)+f^(2)-c)=sqrt(16+25+16)=5`