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  • Equation of the circle described on the chord <img alt="\mathrm{x \cos \alpha+y \sin \alpha=p}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%20%5Ccos%20%5Calpha&plus;y%20%

Equation of the circle described on the chord \mathrm{x \cos \alpha+y \sin \alpha=p} of the circle \mathrm{x^{2}+y^{2}=\mathrm{a}^{2}} as diameter is

Option: 1

\mathrm{x^{2}+y^{2}-a^{2}-2 p(x \cos \alpha+y \sin \alpha-p)=0}


Option: 2

\mathrm{x^{2}+y^{2}+a^{2}+p(x \cos \alpha+y \sin \alpha-p)=0}


Option: 3

\mathrm{x^{2}+y^{2}+a^{2}+2 p(x \cos \alpha+y \sin \alpha-p)=0}


Option: 4

\mathrm{x^{2}+y^{2}-a^{2}+p(x \cos \alpha+y \sin \alpha-p)=0}


Answers (1)

best_answer

Let \mathrm{PQ} be the chord so its equation is:
\mathrm{x \cos \alpha+y \sin \alpha=p}\quad \ldots(1)

Equation of circle through 'PQ' is :
\mathrm{\left(x^{2}+y^{2}-a^{2}\right)+\lambda(x \cos \alpha+y \sin \alpha-p)=0}
\mathrm{x^{2}+y^{2}+\lambda \cos \alpha x+\lambda \sin \alpha y-a^{2}-\lambda p=0}

Its centre is \mathrm{\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)}, It lies


\mathrm{\text { on (1) so }-\frac{\lambda}{2} \cos ^{2} \alpha-\frac{\lambda}{2} \sin ^{2} \alpha=p}
\mathrm{-\frac{\lambda}{2}=\mathrm{p} \text { or } \lambda=-2 \mathrm{p}}

so equation is
\mathrm{x^{2}+y^{2}-a^{2}-2 p(x \cos \alpha+y \sin \alpha-p)=0}

Posted by

Ritika Harsh

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