TS EAMCET · Maths · Circle
The mid-point of the chord of the circle \(x^2+y^2-6 x+4 y-12=0\) drawn parallel to the tangent at \((-1,1)\) and at a distance of one unit from the tangent is
- A \(\left(\frac{3}{4}, \frac{1}{4}\right)\)
- B \(\left(\frac{1}{4}, \frac{3}{4}\right)\)
- C \(\left(\frac{-1}{5}, \frac{2}{5}\right)\)
- D \(\left(\frac{3}{5}, \frac{2}{5}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{-1}{5}, \frac{2}{5}\right)\)
Step-by-step Solution
Detailed explanation
Equation of circle is \(x^2+y^2-6 x+4 y-12=0\) \(\therefore \quad 2 x+2 y y^{\prime}-6+4 y^{\prime}=0\) \(\Rightarrow \quad y^{\prime}=\frac{6-2 x}{2 y+4}\)…
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