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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

Equation of a circle which passes through $(3,6)$ and touches the axes is

$\begin{array}{1 1}x^2+y^2+6x+6y+3=0\\x^2+y^2-6x-6y-9=0\\x^2+y^2-6x-6y+9=0\\x^2+y^2+6x+6y+9=0\end{array} $

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A)
Toolbox:
  • General equation of the circle whose centre is $(h,k)$ is $(x-h)^2+(y-k)^2=a^2$
  • If the circle touches the axes then $h=k=a$=radius of the circle.
Answer : $x^2+y^2-6x-6y+9=0$
It is given that the circle touches the axes of the circle then
$(x-a)^2+(y-a)^2=a^2$
It passes through $(3,6)$
$\therefore (3-a)^2+(6-a)^2=a^2$
On simplifying we get,
$h^2-18h+45=0$
On factorizing we get,
$(h-15)(h-3)=0$
$(h-15)$ is not admissible,
Hence $h=3$
Hence the equation of the circle is $(x-3)^2+(y-3)^2=3^2$
(ie) $x^2+y^2-6x-6y+9=0$
This is the required equation.
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