AP EAMCET · Maths · Circle
A circle touches both the coordinate axes and the straight line \(\mathrm{L} \equiv 4 \mathrm{x}+3 \mathrm{y}-6=0\) in the first quadrant. If this circle lies below the line \(\mathrm{L}=0\), then the equation of that circle is
- A \(4 x^2+4 y^2-4 x-4 y+1=0\)
- B \(4 x^2+4 y^2-4 x-24 y+1=0\)
- C \(x^2+y^2-6 x-6 y+9=0\)
- D \(x^2+y^2-6 x-y-9=0\)
Answer & Solution
Correct Answer
(A) \(4 x^2+4 y^2-4 x-4 y+1=0\)
Step-by-step Solution
Detailed explanation
Center \((h,k)=(r,r)\) \(r = \frac{|4r+3r-6|}{\sqrt{4^2+3^2}}\) \(5r = |7r-6|\) \(7r-6=5r \Rightarrow 2r=6 \Rightarrow r=3\) \(7r-6=-5r \Rightarrow 12r=6 \Rightarrow r=\frac{1}{2}\) For the circle to lie below the line \(4x+3y-6=0\), the center \((r,r)\) must satisfy…
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