Ram Kushwah
Last Activity: 4 Years ago
The eqn of second cicle is:
4(x²+y²)-12x-y-9=0
4x²+4y²-12x-y=9
x² + y² -3x – y/4 = 9/4
(x²-3x+9/4) +(y²-y/4+1/64)=9/4+9/4+1/64
(x-3/2)²+(y-1/8)² =289/64
Or (x-3/2)²+(y-1/8)² =(17/8)²
Thus the centre of this circle is ( 3/2,1/8)
Now the eqn of first cicle is:
x²+y²-3x+λy-5=0
This can be written as
x²-3x+y²+λy=5
{x²-3x+(9/4) }+(y²+λy+λ²/4)=5+9/4+λ²/4
or (x-3/2)² +(y+λ/2)²=(29+λ²)/4
Centre of this cicle is (3/2,-λ/2)
For both the circles to be concentric the
centres should be same
Thus both the following centres are same
(3/2,1/8) and (3/2,-λ/2)
On comparision we get
-λ/2 =1/8
λ=-2/8=-1/4