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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

Equation of the circle described on the chord\mathrm{x \cos \alpha+y \sin \alpha=p}  of the circle \mathrm{x^2+y^2=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 a-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  PQ be  the chord  so its equation is: 

  x cosα + y sinα = p ….(1) 

Equation  of  circle  through ‘PQ’ is

\mathrm{\begin{aligned} & \left(x^2+y^2-a^2\right)+\lambda(x \cos \alpha+y \sin a-p)=0 \\ & x^2+y^2+\lambda \cos \alpha x+\lambda \sin a y-a^2-\lambda p=0 \end{aligned}}

Its centre is \mathrm{\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)} it lies on (1)

\mathrm{-\frac{\lambda}{2} \cos ^2 \alpha-\frac{\lambda}{2} \sin ^2 \alpha=p}

\mathrm{-\frac{\lambda}{2}=p \text { or } \lambda=-2 p}

So the equation is :

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

 

Posted by

avinash.dongre

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