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
- A \(x^2+y^2-4 x+2 y-1=0\)
- B \(x^2+y^2-4 x-2 y=0\)
- C \(x^2+y^2-4 x-2 y+1=0\)
- D \(x^2+y^2-4 x+4 y=0\)
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…
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