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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of the circle which touches the axes and whose centre lies on the line $2x-y=3$?

$\begin{array}{1 1}(a)\;(x-3)^2+(y-3)^2=3^2,(x-1)^2+(y+1)^2=1,(x-3)^2+(y-3)^2=3^2,(x-1)^2+(y+1)^2=1\\(b)\;(x+3)^2+(y+3)^2=3^2,(x+1)^2+(y+1)^2=1,(x+3)^2+(y+3)^2=3^2,(x+1)^2+(y+1)^2=1\\(c)\;(x-3)^2+(y-3)^2=2^2,(x-1)^2+(y+1)^2=0,(x-3)^2+(y-3)^2=2^2,(x-1)^2+(y+1)^2=0\\(d)\;\text{None of these}\end{array}$

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1 Answer

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Since the circle touches both the axis,let the radius of the circle by a,then
Case (i): If centre (a,a) but given centre lies on
$2a-a=3$
$a=3$
Centre =(3,3)
Radius =|3|=3
Equation of circle is $(x-3)^2+(y-3)^2=3^2$
Case (ii): If centre (-a,a) but given centre lies on
$2(-a)-a=3$
$-3a=3$
$a=-1$
Centre =(1,-1)
Radius =|-1|=1
Equation of circle is $(x-1)^2+(y+1)^2=1$
Case (iii): If centre (-a,-a) but given centre lies on
$2(-a)+a=3$
$-a=3$
$a=-3$
Radius =|-3|=3
Equation of circle is $(x-3)^2+(y-3)^2=3^2$
Case (iv): If centre (a,-a) but given centre lies on
$2x-y=3$
$2a+a=3$
$a=1$
Centre =(1,-1),radius =1
Equation of circle is $(x-1)^2+(y+1)^2=1$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
edited Sep 25, 2014 by sharmaaparna1
 

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