logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

If the lines $2x+3y+1=0,\qquad 3x-y-4=0$ lie along diameter of a circle of circumference $10\pi$ then the equation of the circle is :

$\begin{array}{1 1}(A)\;x^2+y^2+2x-2y-23=0\\(B)\;x^2+y^2-2x-2y-23=0 \\(C)\;x^2+y^2+2x+2y-23=0 \\(D)\;x^2+y^2-2x+2y-23=0 \end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Two diameters are along
$2x+3y+1=0$
$3x-y-4=0$
on solving we get centre as $(1,-1)$
We know that circumference $=2 \pi r$
$2 \pi r=10 \pi$
$2r= 10$
$r=5$
Required circle is $(x-1)^2+(y+1)^2=5^2$
$x^2+y^2-2x+2y-23=0 $
Hence D is the correct answer.
answered Apr 8, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...