TS EAMCET · Maths · Circle
The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines \(x=0\) and \(y=0\) is
- A \(x^2+y^2-6 x+6 y+9=0\)
- B \(x^2+y^2-6 x-6 y+9=0\)
- C \(x^2+y^2+6 x-6 y+9=0\)
- D \(x^2+y^2+6 x+6 y+9=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-6 x+6 y+9=0\)
Step-by-step Solution
Detailed explanation
Given, radius \(=3\), Since, the circle touching both the coordinate axes in 4th quadrant, so equation is \(\begin{gathered}\quad(x-3)^2+(y+3)^2=3^2 \\ \Rightarrow \quad x^2+9-6 x+y^2+9+6 y=9 \\ \Rightarrow \quad x^2+y^2-6 x+6 y+9=0\end{gathered}\)
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