Please answer the question in detail and fast and no linking.Solve10th question

Please answer the question in detail and fast and no linking.Solve10th question ball is moving around the circle 14x2 +14y2 clockwise direction leaves tangentially at the point 6) after getting reflected from a straight line L = O it passes through the centre of 11 the circle. The perpendicular distance of this straight line L O from the point P is — v/fiö . You can assume 13 that the angle of incidence is equal to the angle of reflection. 10. 11. 12. The equation of tangent to the circle at P is (A) 2x-Y+ 12=0 -2y + 21 = o Radius of the circle is 165 14 165 (C) (B) 4x• 3y-6=O (D)2x+ -O 165 46 (D) None of these If angle between the tangent at P and the line through 'P' perpendicular to the line L = O is 0, then tan O is (A) 2/11 (C) 4/11 (B) 3/11 (D) None of these

We have, 14 x^{2} + 14 y^{2} + 216 x - 69 y + 432 = 0 \Rightarrow x^{2} + y^{2} + \frac{108}{7} x - \frac{69}{14} y + \frac{216}{7} = 0 \Rightarrow x^{2} + \frac{2916}{49} + \frac{108}{7} x + y^{2} + \frac{4761}{784} - \frac{69}{14} y - \frac{2916}{49} - \frac{4761}{784} + \frac{216}{7} = 0 \Rightarrow x^{2} + \frac{2916}{49} + \frac{108}{7} x + y^{2} + \frac{4761}{784} - \frac{69}{14} y - \frac{27225}{784} = 0 \Rightarrow {\{x + \frac{54}{7}\}}^{2} + {\{y - \frac{69}{28}\}}^{2} = {(\frac{165}{28})}^{2} So, Centre of circle is (- \frac{54}{7}, \frac{69}{28}) and radius of circle = \frac{165}{28} Slope, m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - \frac{69}{28}}{- 3 - (- \frac{54}{7})} = \frac{\frac{99}{28}}{\frac{33}{7}} = \frac{3}{4} Equation of tangent at P (- 3, 6) is y - y_{2} = - \frac{1}{m} (x - x_{2}) \Rightarrow y - 6 = - \frac{1}{\frac{3}{4}} (x + 3) \Rightarrow y - 6 = \frac{- 4}{3} (x + 3) \Rightarrow 3 y - 18 = - 4 x - 12 \Rightarrow 4 x + 3 y - 6 = 0

View Full Answer