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  • A line is drawn through a fixed point P (α, β) to cut the circle <img alt="x^{2}+y^{2}=r^{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B2%7D&

A line is drawn through a fixed point P (α, β) to cut the circle x^{2}+y^{2}=r^{2}at A and B. Determine PA ⋅ PB in terms of α, β and r.

 

Option: 1

\sqrt{\alpha ^{2}+\beta ^{2}-\gamma ^{2}}


Option: 2

(\alpha ^{2}+\beta ^{2})\cdot \gamma ^{2}


Option: 3

\alpha +\beta -\gamma ^{2}


Option: 4

\alpha ^{2}+\beta ^{2}-\gamma ^{2}


Answers (1)

best_answer

Any line through (α, β) is \mathrm{\frac{x-\alpha}{\cos \theta}-\frac{y-\beta}{\sin \theta}=k}

Any point on it is (α+ k cos θ, β + k sin θ) 

This point will lie on the given circle if 

\mathrm{\begin{aligned} & (\alpha+k \cos \theta)^2+(\beta+k \sin \theta)^2=r^2 \\ & \text { i.e. } k^2+2 k[\alpha \cos \theta+\beta \sin \theta]+\left[\alpha^2+\beta^2-r^2\right]=0 \\ & \text { Let PA }=k_1, P B=k_2 \end{aligned}}-------(1)

∴ k1, k2 are the roots of (1)

\mathrm{\Rightarrow \mathrm{PA} \cdot \mathrm{PB}=\mathrm{k}_1 \cdot \mathrm{k}_2=\alpha^2+\beta^2-r^2}.

 

 

Posted by

avinash.dongre

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