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COMEDK · Maths · 12. Circle

The circle \(x^{2}+y^{2}-8 x+4 y+4=0\) touches

  1. A \(X\)-axis
  2. B \(Y\)-axis
  3. C both axes
  4. D neither \(X\)-axis nor \(Y\)-axis.
Verified Solution

Answer & Solution

Correct Answer

(B) \(Y\)-axis

Step-by-step Solution

Detailed explanation

We have, \(x^{2}+y^{2}-8 x+4 y+4=0\) \(\Rightarrow\left(x^{2}-8 x+16\right)+\left(y^{2}+4 y+4\right)=16\) \[ \Rightarrow \quad(x-4)^{2}+(y+2)^{2}=4^{2} \] \(\therefore\) Centre is \((4,-2)\) and radius is 4 . Hence, given circle touches \(Y\)-axis.