TS EAMCET · Maths · Circle
The circle \(S=0\) cuts the circles \(C_1=x^2+y^2-8 x-2 y+16=0\) and \(C_2=x^2+y^2-4 x-4 y-1=0\) orthogonally. If the common chord of \(S=0\) and \(C_1=0\) is \(2 x+13 y-15=0\), then the centre of \(S=0\) is
- A \(\left(\frac{-11}{3}, \frac{7}{6}\right)\)
- B \(\left(\frac{11}{3}, \frac{-7}{6}\right)\)
- C \(\left(\frac{2}{13}, \frac{11}{15}\right)\)
- D \(\left(\frac{11}{15}, \frac{-2}{13}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{-11}{3}, \frac{7}{6}\right)\)
Step-by-step Solution
Detailed explanation
Let the equation of circle \(S\) is \(\begin{aligned} & S=x^2+y^2+2 g x+2 f y+c=0 \\ & C_1=x^2+y^2-8 x-2 y+16=0 \\ & C_2=x^2+y^2-4 x-4 y-1=0 \end{aligned}\) \(S\) is orthognoally to \(C_1\) and \(C_2\) From Eqs. (i) and (ii), Common chord of \(S\) and \(C_1=\) Given common chord…
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