If `-3l^2-6l-1+6m^2=0`, Then the equation 0 is a If of the circle for which `lx + my +1=0` tangent, is

If 7l^2 - 9m^2 + 8l + 1 = 0 and we have to find equation of circle having lx + my + 1 = 0 is a tangent and we can adjust given condition as 16l^2 + 8l + 1 = 9(l^2 + m^2) or (4l^2 + 1)^2 = 9(l^2 + m^2) implies |4l + 1|/sqrt ((l^2 + m^2)) = 3 Centre of circle = (4, 0) and radius = 3 when any two non parallel lines touching a circle, then centre of circle lies on angle bisector of lines. Answer the following question based on above passage : If 16m^2 - 8l - 1 = 0 , then equation of the circle having lx + my + 1 = 0 is a tangent is

If 4t^(2)-5m(2)+6l+1=0 , then the line lx+my+1=0 touches the circle

If 16l^(2)+9m^(2)=24lm+6l+8m+1 and the line mx+ly=1 is tangent to circle S_(1)=0 then the equation of circle S_(2)=0 which touches S_(1)=0 at (0,0) and passing (3,1) is given by x^(2)+y^(2)-ax-by+c=0, then find the value of sqrt(3a+4b+5c)

A line is tangent to a circle if the length of perpendicular from the centre of the circle to the line is equal to the radius of the circle. If 4l^2 - 5m^2 + 6l + 1 = 0 , then the line lx + my + 1=0 touches a fixed circle whose centre. (A) Lies on x-axis (B) lies on yl-axis (C) is origin (D) none of these

A line is tangent to a circle if the length of perpendicular from the centre of the circle to the line is equal to the radius of the circle. If 4l^2 - 5m^2 + 6l + 1 = 0 , then the line lx + my + 1=0 touches a fixed circle whose centre. (A) Lies on x-axis (B) lies on yl-axis (C) is origin (D) none of these

A line is tangent to a circle if the length of perpendicular from the centre of the circle to the line is equal to the radius of the circle. If 4l^2 - 5m^2 + 6l + 1 = 0 , then the line lx + my + 1=0 touches a fixed circle whose centre. (A) Lies on x-axis (B) lies on yl-axis (C) is origin (D) none of these