P(a,b),Q(c,d) & R(e,f) are three points satisfying the inequality x2+y2−6x−8y<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 PQR intersects the tangents to the circle x2+y2−6x−8y=0 drawn at origin and (c+e2+1,b). If area of the triangle formed by these three lines (two tangents and internal bisector of angle P) is Δ, then the value of 3Δ100 is ___

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$P (a, b), Q (c, d) & R (e, f)$ are three points satisfying the inequality $x^{2} + y^{2} - 6 x - 8 y < 0$ , where $a, b, c, d, e, & f \in I$ , 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 PQR intersects the tangents to the circle $x^{2} + y^{2} - 6 x - 8 y = 0$ drawn at origin and $(\frac{c + e}{2} + 1, b)$ . If area of the triangle formed by these three lines (two tangents and internal bisector of angle P) is $Δ$ , then the value of $\frac{3 Δ}{100}$ is ____.

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Detailed Solution

$x^{2} + y^{2} - 6 x - 8 y < 0 {(x - 3)}^{2} + {(y - 4)}^{2} - 25 < 0$

Point atleast distance from $(- 2, 4) i s P (a, b) \equiv P (- 1, 4)$
Points which are greatest distance from $(- 2, 4) a r e Q (c, d)$ & $R (e, f) \equiv Q (7, 6) & R (7, 2)$
$Δ P Q R$ is an isosceles triangle & internal bisector of $∠ P i s y = 4$
Equation of tangent at origin is $3 x + 4 y = 0$
Equation of tangent at $(\frac{c + e}{2} + 1, b) \equiv (8, 4)$ is x=8
Area of the right-angled triangle formed by above three lines is $ϕ = \frac{1}{2} \times 10 \times \frac{40}{3} = \frac{200}{3}$