AP EAMCET · Maths · Circle
If \((3,-1)\) is one end of a diameter of the circle \(x^2+y^2-2 x+4 y=0\), then the equation of the tangent at the other end of that diameter is
- A \(2 x+y-5=0\)
- B \(2 x+y+5=0\)
- C \(x+2 y+5=0\)
- D \(x+2 y-5=0\)
Answer & Solution
Correct Answer
(B) \(2 x+y+5=0\)
Step-by-step Solution
Detailed explanation
Center of circle: \(C = (1, -2)\) Other end of diameter \( (x_2, y_2) \): \(\frac{3+x_2}{2}=1 \Rightarrow x_2=-1\), \(\frac{-1+y_2}{2}=-2 \Rightarrow y_2=-3\) Other end: \((-1, -3)\) Tangent at \((x_0, y_0)\) on \(x^2+y^2+2gx+2fy+c=0\) is \(x x_0+y y_0+g(x+x_0)+f(y+y_0)+c=0\)…
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