Find the equation of a circle circumscribing the triangle whose side are x=0,y=0 and lx+my=1.if l,m vary so that l^2+m^2=4l^2m^2,find the locus of the circumcentre of the circle?

#l^2+m^2=4l^2m^2#

1 Answer
Jul 14, 2018

# (1) : x^2+y^2-x/l-y/m=0#.

# (2) : x^2+y^2=1#.

Explanation:

Let #l_1 : x=0, l_2 : y=0, l_3 : lx+my=1# be the sides of the #Delta#.

Clearly, the vertices of the #Delta# are given by,

# l_1nnl_2={O(0,0)}#,

#l_2nnl_3={B(1/l,0)}#

#l_3nnl_1={A(0,1/m)}#.

Let us note that, #/_AOB=90^@#.

#:. AB# is a diameter of the circumcircle of #Delta ABC#.

We know that, the eqn. of a circle having diametric

extremeties #(x_1,y_1) and (x_2,y_2)# is

#(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0#.

So, the eqn. of the reqd. circle is,

#(x-0)(x-1/l)+(y-1/m)(y-0)=0#,

# i.e., x^2+y^2-x/l-y/m=0#.

Now #AB# is a diameter, so, its mid-point is the centre.

Hence, the centre is #(1/(2l),1/(2m))#.

To find the locus of the centre, let, #(X,Y)=(1/(2l),1/(2m))#.

#l^2+m^2=4l^2m^2..."[Given]"#

#rArr l^2/(4l^2m^2)+m^2/(4l^2m^2)=1#.

#rArr 1/(2m)^2+1/(2l)^2=1#.

#rArr X^2+Y^2=1#.

So, the locus of the circumcentre is #x^2+y^2=1#, which

is the unit circle.

#color(green)("Enjoy Maths.!")#