Question

A Panchayath has a circular park of diameter 13 metres (Figure 1). Panchayath committee
decided to renovate it. They erected a pole (P) on the boundary of a circular park in such
a way that the differences of its distances from two diametrically opposite fixed gates A
and B is 7 metres (Figure2).
B В
13
P
Figure 1
Figure 2
a) if BP =x metre expression for ap. b) equation relating to ap, bp and ab in terms of
c) quadratic equation in x is
d) discriminant of x^2 + 7x-60=0
e) nature of roots of x^2 + 7x-60=0
​

Answer 1

Step-by-step explanation:

its the correct answer. thank you

Answer 2

Step-by-step explanation:

According to theorem related to circles, any point on the circle form right angle with end points of the diameter of the circle. Let the distance of the pole point P from the A is x and the distance from the point B is y. (End points of the diameter AB  ).

Without loss of generality, assume y>x. then according to the given condition

$y-x = 7 m$.......(1)

and by baudhayan(pythagorous) theorem, we have

$x^2+y^2=13^2=169$.....(2)

Solving equation (1) and (2) we get,

$x^2 + (x+7)^2 =169\\\implies x^2+7x-60=0\\Discriminant = \sqrt{(7)^2-4(1)(-60)}=17\\ Roots \,\, given \,\, by \\(x+12)(x-5)=0\\\implies x= -12,5$

Nature of the roots is real and distinct.

since distance cannot be negative hence

x= 5m and y= 12m