The line $x-2=0$ cuts the circle $x^2+y^2-8 x-2 y+8=0$ at $A$ and $B$. The equation of the circle passing…

The line $x-2=0$ cuts the circle $x^2+y^2-8 x-2 y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having least radius is
  1. $x^2+y^2-4 x+2 y-1=0$
  2. $x^2+y^2-4 x-2 y=0$
  3. $x^2+y^2-4 x-2 y+1=0$
  4. $x^2+y^2-4 x+4 y=0$

Solution

Equation of circles passes through the point of intersection of line $x-2=0$ and the circle $x^2+y^2-8 x-2 y+8=0$, is $ \left(x^2+y^2-8 x-2 y+8\right)+\lambda(x-2)=0 $
For minimum radius, it is necessary that centre of circle Eq. (i) lies on the line $x-2=0$, so $ \begin{gathered} -\left(\frac{\lambda-8}{2}\right)-2=0 \\ \Rightarrow \quad \lambda-8=-4 \quad \Rightarrow \lambda=4 \end{gathered} $ So, equation of required circle is $ x^2+y^2-4 x-2 y=0 $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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