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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

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

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

1 Answer

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
 

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