Let #l_1 : 4x-3y+10=0, and, l_2 : 3x+4y-30=0# be the
given tgt. lines to the reqd. circle, say, #S#. Note taht, #l_1 bot l_2#.
Also, let #P(-1,2) and Q(6,3)# be the pts. of contact of #S# with
#l_1 and l_2#, resp.
Let pt. #C# be the Centre, and, #r# the radius, of #S#. Then, we
know, from Geometry, that,
#"dist."CP="dist."CQ =r, and, "line "CP bot l_1, "line "CQ bot l_2#.
Since, #"line CP"nn"line "CQ={C}#, we will obtain the co-ords. of
the centre #C# by solving the eqns. of these lines.
Eqn. of line CP :-
#"Line "CP botl_1, &, l_2botl_1#
#rArr " line "CP |\| l_2 :3x+4y-30=0," with "P in CP#
#" Eqn. of "CP : 3x+4y=3(-1)+4(2)=5.........(1)#
Similarly, Eqn. of line CQ #: 4x-3y=15..................(2)#
Solving #(1), &, (2)#, we get, the Centre #C(3,-1)#.
Then, #r=dist. CP=sqrt{(3+1)^2+(2+1)^2}=5#
Finally, the eqn. of #S : (x-3)^2+(y+1)^2=5^2#, or,
#x^2+y^2-6x+2y-15=0#
Enjoy Maths.!
