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

Find the equation of circle passing through (0,0) and (0,B)?

$\begin{array}{1 1}(a)\;(x-\large\frac{B}{2})^2\normalsize+(y-\large\frac{B}{2})^2=\normalsize(\large\frac{B}{\sqrt 2})^2\\(b)\;(x-\large\frac{B}{2})\normalsize+(y-\large\frac{B}{2})=\normalsize(\large\frac{B}{\sqrt 3})^2\\(c)\;(x+\large\frac{B}{2})^2\normalsize+(y+\large\frac{B}{2})^2=\normalsize(\large\frac{B}{\sqrt 2})^2\\(d)\;\text{None of these}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Let $(h,k)$ be the centre of circle then the circle passing through (0,0).
OA=AB=AC=radius
By similarity we can say
OB=OC
Hence if B(0,B) then C point will be (B,0) hence three points which are passing through circle are (0,0),(B,0),(0,B) in which (B,0) and (0,B) are diametrically opposite.
Hence centre (h,k) will be
$h=\large\frac{B}{2},$$k=\large\frac{B}{2}$
Radius=$\large\frac{B}{\sqrt 2}$
Circle equation =$(x-\large\frac{B}{2})^2+$$(y-\large\frac{B}{2})^2=(\large\frac{B}{\sqrt 2})^2$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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
...