Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A line is drawn through a fixed point to cut the circle<img alt="\mathrm{x^2+y^2=

A line is drawn through a fixed point \mathrm{P(h, k)} to cut the circle\mathrm{x^2+y^2=a^2 \: at \: Q\: and\: R}. Then \mathrm{P Q. P R} is equal to
 

Option: 1

\mathrm{(h+k)^2-a^2}

 


Option: 2

\mathrm{h^2+k^2-a^2}
 


Option: 3

\mathrm{(h-k)^2+a^2}
 


Option: 4

\mathrm{h^2+k^2+a^2}


Answers (1)

Any line through \mathrm{P(h, k)\: is\: \frac{x-h}{\cos \theta}=\frac{y-k}{\sin \theta}=r}...........(i)

Any point on (i) is \mathrm{(r \cos \theta+h, r \sin \theta+k)}

Line (i) cuts the circle \mathrm{x^2+y^2=a^2}      ...........(ii)

\mathrm{\Rightarrow (r \cos \theta+h)^2+(r \sin \theta+k)^2=a^2 }

\mathrm{\Rightarrow r^2+2 r(h \cos \theta+k \sin \theta)+\left(h^2+k^2-a^2\right)=0}...........(iii)

Line (i) cuts circle at \mathrm{Q \: and \: R}

Let \mathrm{P Q=r_1, P R=r_2}

Then \mathrm{r_1, r_2} are roots of (iii)

\mathrm{ \Rightarrow \quad r_1 r_2=\left(h^2+k^2-a^2\right) }

Hence option 2 is correct.

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question