AP EAMCET · Maths · Straight Lines
If \(P\) is a variable point which is at a distance of 2 units from the line \(2 x-3 y+1=0\) and \(\sqrt{13}\) units from the point \((5,6)\), then the equation of the locus of P is
- A \(4 x^2+12 x y-5 y^2-44 x-42 y+245=0\)
- B \(12 x y-5 y^2-44 x-42 y+243=0\)
- C \(8 x^2+12 x y-5 y^2-44 x-42 y+243=0\)
- D \(12 x y-13 y^2-44 x-42 y+245=0\)
Answer & Solution
Correct Answer
(B) \(12 x y-5 y^2-44 x-42 y+243=0\)
Step-by-step Solution
Detailed explanation
\( \frac{|2x-3y+1|}{\sqrt{2^2+(-3)^2}} = 2 \) \( (2x-3y+1)^2 = (2\sqrt{13})^2 \) \( 4x^2 - 12xy + 9y^2 + 4x - 6y + 1 = 52 \) \( E_1: 4x^2 - 12xy + 9y^2 + 4x - 6y - 51 = 0 \) \( \sqrt{(x-5)^2 + (y-6)^2} = \sqrt{13} \) \( (x-5)^2 + (y-6)^2 = 13 \)…
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