TS EAMCET · Maths · Circle
If a variable circle \(S=0\) touches the line \(y=x\) and passes through the point \((0,0)\), then the fixed point that lies on the common chord of the circles \(x^2+y^2+6 x+8 y-7=0\) and \(S=0\) is
- A \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
- B \(\left(-\frac{1}{2},-\frac{1}{2}\right)\)
- C \(\left(\frac{1}{2},-\frac{1}{2}\right)\)
- D \(\left(-\frac{1}{2}, \frac{1}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{1}{2}, \frac{1}{2}\right)\)
Step-by-step Solution
Detailed explanation
Let the equation of circle \(S=0\), passes through the point \((0,0)\) is \(S=x^2+y^2+2 g x+2 f y=0\) Since, circle \(S=0\) touches the line \(y=x\), so…
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