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प्रश्न
Find the angle of intersection of the following curve x2 + y2 − 4x − 1 = 0 and x2 + y2 − 2y − 9 = 0 ?
उत्तर
\[\text{ Given curves are },\]
\[ x^2 + y^2 - 4x - 1 = 0 . . . \left( 1 \right)\]
\[ x^2 + y^2 - 2y - 9 = 0 . . . \left( 2 \right)\]
\[\text { From } (3)\text { we get }\]
\[ x^2 + y^2 = 4x + 1\]
\[\text { Substituting this in} (2),\]
\[4x + 1 - 2y - 9 = 0\]
\[ \Rightarrow 4x - 2y = 8\]
\[ \Rightarrow 2x - y = 4\]
\[ \Rightarrow y = 2x - 4 . . . \left( 3 \right)\]
\[\text { Substituting this in } (1),\]
\[ x^2 + \left( 2x - 4 \right)^2 - 4x - 1 = 0\]
\[ \Rightarrow x^2 + 4 x^2 + 16 - 16x - 4x - 1 = 0\]
\[ \Rightarrow 5 x^2 - 20x + 15 = 0\]
\[ \Rightarrow x^2 - 4x + 3 = 0\]
\[ \Rightarrow \left( x - 3 \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow x = 3 orx = 1\]
\[\text { Substituting the values of } x in \left( 3 \right), \text { we get,} \]
\[y = 2 or y = - 2 \]
\[ \therefore \left( x, y \right)=\left( 3, 2 \right),\left( 1, - 2 \right)\]
\[\text { Differentiating (1) w.r.t.x },\]
\[2x + 2y \frac{dy}{dx} - 4 = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{4 - 2x}{2y} = \frac{2 - x}{y} . . . \left( 4 \right)\]
\[\text { Differenntiating (2) w.r.t.x },\]
\[2x + 2y \frac{dy}{dx} - 2\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx}\left( 2y - 2 \right) = - 2x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2x}{2 - 2y} = \frac{x}{1 - y} . . . \left( 5 \right)\]
\[\text { Case }- 1:\left( x, y \right)=\left( 3, 2 \right)\]
\[\text { From } \left( 4 \right), \text { we get }, m_1 = \frac{2 - 3}{2} = \frac{- 1}{2}\]
\[\text { From } \left( 5 \right), \text { we get }, m_2 = \frac{3}{1 - 2} = - 3\]
\[\text { Now }, \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{- 1}{2} + 3}{1 + \frac{3}{2}} \right| = 1\]
\[ \Rightarrow \theta = \tan^{- 1} \left( 1 \right) = \frac{\pi}{4}\]
\[\text { Case-}2: \left( x, y \right)=\left( 1, - 2 \right)\]
\[\text { From } \left( 4 \right), \text { we get,} m_1 = \frac{2 - 1}{- 2} = \frac{- 1}{2}\]
\[\text { From } \left( 5 \right), \text { we get }, m_2 = \frac{1}{1 + 2} = \frac{1}{3}\]
\[\text { Now,} \]
\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{- 1}{2} - \frac{1}{3}}{1 - \frac{1}{6}} \right| = 1\]
\[ \Rightarrow \theta = \tan^{- 1} \left( 1 \right) = \frac{\pi}{4}\]
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