P ( a , b ), Q ( c , d ) & R ( e , f) are three points satisfying the inequality x ^2+ y ^2-6 x -8 y 0 , where a , b , c , d , e , & f I , such that P is situated at least distance while Q & R are situated at greatest distance from the point A(-2,4) . Now internal bisector of angle P of triangle P Q R intersects the tangents to the circle x^2+y^2-6 x-8 y=0 drawn at origin and ( c+e 2 +1, b ) . If area of the triangle

Step-by-step solution

aligned & x^2+y^2-6 x-8 y 0 & (x-3)^2+(y-4)^2-25 0 aligned Point atleast distance from (-2,4) is P ( a , ~b ) P (-1,4) Points which are greatest distance from (-2,4) are Q ( c , d ) & R ( e , f) Q (7,6) & R (7,2) PQR is an isosceles triangle & internal bisector of P is y =4 Equation of tangent at origin is 3 x+4 y=0 equation of tangent at ( c+e 2 +1, b ) (8,4) is x =8 Area of the right angled triangle formed by above three lines is = 1 2 10 40 3 = 200 3 3 =200