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

Find the equation of the circle having $(1,-2)$ as its centre and passing through $3x+y=14,2x+5y=18$.

$\begin{array}{1 1}x^2+y^2+2x+4y+20=0\\x^2+y^2-2x+4y+20=0\\x^2+y^2-2x-4y+20=0\\x^2+y^2+2x-4y+20=0\end{array} $

The answer is (2)

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A)
Toolbox:
  • General equation of the circle is $x^2+y^2+2hx+2ky+c=0$
Answer : $x^2+y^2+2x-4y+20=0$
Let the equation of the given circle be $x^2+y^2+2hx+2ky+c=0$
Since the coordinates of the centre is given as $(1,-2)$,the equation of the circle is (ie) $h=1$ and $k=-2$.
$x^2+y^2+2x-4y+c=0$
Let us now find the point of intersection of the two lines
$3x+y=14$ and $2x+5y=18$
$(\times 2)3x+\;y=14$
$(\times 3)2x+5y=18$
____________________
$6x+\;2y=28$
$6x+15y=54$
_____________________
$-13y=-26$
$\Rightarrow y =2$
$\therefore 3x+(2)=14$
$\Rightarrow x=4$
Hence the point of intersection is $(4,2)$
Since this point lies on the circle
$(4)^2+(2)^2+2(4)-4(2)+c=0$
$\Rightarrow c=20$
Hence the equation of the required circle is $x^2+y^2+2x-4y+20=0$
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