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AP EAMCET · Maths · Circle

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. A \(x^2+y^2-4 x+2 y-1=0\)
  2. B \(x^2+y^2-4 x-2 y=0\)
  3. C \(x^2+y^2-4 x-2 y+1=0\)
  4. D \(x^2+y^2-4 x+4 y=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(x^2+y^2-4 x-2 y=0\)

Step-by-step Solution

Detailed explanation

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…