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`3sqrt2``2sqrt2`23

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Since,the equation of family of circles touching line <br> L=0 a t their point of contact `(x_(1),y_(1))` " is " `(x-x_(1))^(2) +(y-y_(1))^(2)+lambdaL=0, " where " Lambda in R`. <br> `therefore` Equation of circle, touches the x= y at point (1, 1) is <br> `(x-1)^(2)+(y-1)^(2)+lambda(x-y)=0` <br> `rArr x^(2)+y^(2)+(lambda-2)x+(-lambda-2)y+2=0" "...(i)` <br> ` therefore` CIrcle(i) passes through point (1, -3). <br> `therefore 1 + 9 +(lambda-2)+3(lambda+2)+2=0` <br> `rArr 4lambda + 16 = 0` <br> `rArr lambda = - 4` <br> So, equation of circle (i) at `lambda =-4` , is <br> `x^(2)+y^(2)-6x+2y+2=0` <br> Now, radius of the circle `=sqrt(9+1-2)=2sqrt2`.