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Let the abscissae of the two points P and Q be the roots of 2x^{2}-rx+p=0 and the ordinates of P and Q be the roots of x^{2}-sx-q=0. If the equation of the circle described on PQ as diameter is  2\left(x^{2}+y^{2}\right)-11 x-14 y-22=0, \text { then } 2 \mathrm{r}+\mathrm{s}-2 \mathrm{q}+\mathrm{p} is equal to ___________.

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let co-ordinate of P and Q are \mathrm{\left ( \alpha ,\beta \right )} and \mathrm{\left ( \gamma ,\delta \right )} respectively
\mathrm{\alpha+\gamma=\frac{r}{2}, \alpha \gamma=\frac{p}{2}}
\mathrm{and \, \beta+\delta=s, \beta \delta=-q}

Equation of circle with \mathrm{PQ} as diameter
\mathrm{(x-\alpha)(x-\gamma)+(y-\beta)(y-\delta)=0}
\mathrm{x^{2}+y^{2}-x(\alpha+\gamma)-y(\beta+\delta)+\alpha \gamma+\beta \delta=0}
\mathrm{x^{2}+y^{2}-x\left(\frac{r}{2}\right)-y(S)+\frac{p}{2}-q=0\, ---(I)}

\mathrm{S: 2\left(x^{2}+y^{2}\right)-11 x-14 y-22=0}
\mathrm{S: x^{2}+y^{2}-\frac{11}{2} x-7 y-11=0\; ----(II)}

\mathrm{\because \; e q (I)\: \&\: (II) represent\; same \; circle}
\mathrm{\left.\begin{matrix} \therefore \frac{r}{2}= \frac{11}{2}\\ r= 11 \end{matrix}\right|\left.\begin{matrix} S= 7\\ \end{matrix}\right|\begin{matrix} \frac{P}{2}-q= -11\\ \frac{P-2q}{2}= -11 \\ P-2q= -22 \end{matrix}}

\mathrm{\therefore \; 2r+S-2q+P}
       \mathrm{2\times 11+7-22= 7}

Posted by

HARSH KANKARIA

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