Question

. If x - y + 1 = 0 meets the circle x^2 + y^2 + y-1 = 0 at A and B then the equation of the circle with AB as
diameter is​

Answer 1

⠀$~~~~~~~~~~~~~~~~~~~\sf\pink{S+λL=0}, λ\: being\: a\: constant.$

$~~~~~~~~~~~~~~~~~~~\sf{x^ 2 +y ^2 +y−1+λ(x−y+1)\:=\:0}$

⠀$~~~~~~~~~~~~~~~~~~~~~~~\red{\sf{ Centre=}}\sf (\frac{-λ}{2},\frac{λ-1}{2} )$

⠀$~~~~~~~~~~~~~~~~~~\red{\sf{Centre \:lies\: on\:the\: line:}} \sf {x−y+1=0}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf\red{⇒}{λ= \frac{3}{2}}$

$\red{\sf{ Equation \:thus\; becomes :}}\sf{\:x^2 +y ^2+y−1+ \frac{3}{2} (x−y+1)=0}$

⠀⠀$~~~~~~~~~~~~~~~~~~ \red{\sf{i.e.,\:}} \sf{ 2( x^2+y^2)+3x−y+1 = 0}$

$\sf {Therefore, the\: equation \:of\: the\: circle\: with\: AB \:as\: diameter\: is\:}$

${ 2( x^2+y^2)+3x−y+1 = 0}}$