If `4l^2-5m^2+6l+1=0.` Prove that `lx+my+1=0` touches a definite circle. Find the centre & radius of the circle.

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

The equation of the circle with centre (6,0) and radius 6 is

If 8l^2+35 m^2+36 l m+6l+12 m+1=0 and the line l x+m y+1=0 touches a fixed circle whose centre is (alpha,beta) and radius is ' r^(prime), then the value of (alpha+beta-r) is equal to

A circle centre (1,2) touches y-axis. Radius of the circle is

Find the equation of the circle with centre :(0, 2) and radius 2

Find the equation of the circle with: centre (0,-1) and radius 1.

Find the equation of the circle with centre at (0, 0) and radius 4.

Consider the relation 4l^(2)-5m^(2)+6l+1=0 , where l, m inR . The line lx+my+1=0 touches a fixed circle whose equation is

If the line l x+m y-1=0 touches the circle x^2+y^2=a^2 , then prove that (l , m) lies on a circle.