(a)OR(b)

There is a circular park of diameter $\displaystyle 65$ m as shown in the following figure, where AB is a diameter.

An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is $\displaystyle 35$ m more than the distance of P from B . Find distance of point P from A and B respectively.

Let distance of gate at P from point B is \(\displaystyle x\) m Then distance of gate at P from point A is \(\displaystyle (35+x) \mathrm{m}\) In right \(\displaystyle \Delta \mathrm{APB}\) \[\begin{aligned} & (x+35)^{2}+x^{2}=(65)^{2} \\ & x^{2}+35 x-1500=0 \\ & (x+60)(x-25)=0 \\ & x=25 \end{aligned} \] Hence, \(\displaystyle x+35=60\) Distance of P from A = $\displaystyle 60$ m Distance of P from B = $\displaystyle 25$ mFor real roots, \(\displaystyle \mathrm{D} \geq 0\) \[\begin{aligned} {[-2(\mathrm{p}+1)]^{2}-4 \mathrm{p}^{2} } & \geq 0 \\ \Rightarrow \mathrm{p} & \geq-\frac{1}{2} \end{aligned} \] ∴ smallest value of \(\displaystyle \mathrm{p}=-\frac{1}{2}\) At \(\displaystyle \mathrm{p}=-\frac{1}{2}\) given equation becomes \[x^{2}-2\left(\frac{-1}{2}+1\right) x+\left(\frac{-1}{2}\right)^{2}=0 \] \(\displaystyle x^{2}-x+\frac{1}{4}=0\) or \(\displaystyle 4 x^{2}-4 x+1=0\) \[(2 x-1)(2 x-1)=0 \] ∴ roots are \(\displaystyle \frac{1}{2}, \frac{1}{2}\)