TS EAMCET · Maths · Circle
Let 6,8 be the \(\mathrm{X}\) and \(\mathrm{Y}\) - intercepts made by the circle \(\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) respectively. If \(\mathrm{gx}+\mathrm{fy}+1\) \(=0\) is a line passing through the point \((1,-1)\), then the radius of the circle \(S=0\) is
- A \(\sqrt{41}\)
- B \(13\)
- C \(\sqrt{26}\)
- D \(5\)
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
\(2 \sqrt{g^2-c}=6 \Rightarrow g^2-c=9\) ...(1) \(2 \sqrt{f^2-c}=8 \Rightarrow f^2-c=16\) ...(2) \(g x+f y+1=0\) passes through \((1,-1)\) \(\therefore g-f+1=0 \Rightarrow g-f=-1\) ...(3) Equations (1) - (2), \(g^2-f^2=-7\) \(\Rightarrow(g+f)(g-f)=-7 \Rightarrow g+f=7\) ...(4)…
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