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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 :-A. 5B. `2sqrt5`C. `sqrt57`D. 4 |
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Answer» Correct Answer - A 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` |
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