Shrutik
Last Activity: 6 Years ago
Let the centre of the circle be C (h,k).
Since the Centre lies on the line 4x+3y=2, we get 4h+3k=2
h = ( 2 – 3k ) / 4 …... Eqn (1)
Since the lines 7x-y+4=0 and x+y+4 are touching the circles ( ie they are tangents ), let’s use the formula : Distance of point C(h,k) from these lines to get the radius as :
…...Eqn (2)
Substitute Eqn 1 in Eqn 2 to solve for k, giving k = -2 … thus h = 2 for +ve sign of Modulus and k = 6 and h = -4 for -ve sign of modulus.
Thus, the 2 centres are C1(2,-2) and C2(-4,6). Radii of these 2 circles are 4/sqrt(2) and 6/sqrt(2) respectively.
Finally, using Centre -radius form, we get the 2 equations as
x2 + y2 -4x+4y = 0 and x2+y2+8x-12y+34=0
