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What is the centre and radius of the circle with the equation x^{2}+y^{2}-4x+6y-11= 0 ?

Option: 1

Centre: \left ( 2,-3 \right ),Radius : 2\sqrt{6}


Option: 2

Centre: \left ( -2,3 \right ),Radius : 2\sqrt{6}


Option: 3

Centre:\left ( 2,-3 \right ) ,Radius : 3\sqrt{6}


Option: 4

Centre: \left ( -2,3 \right ),Radius : 3\sqrt{6}


Answers (1)

best_answer

To find the centre and radius of the given circle equation, we first need to write it in the standard form \left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}. We can do this by completing the square as follows:
x^{2}+y^{2}-4x+6y-11= 0
\left ( x^{2}-4x \right )+\left ( y^{2}+6y \right )= 11
\left ( x^{2}-4x+4 \right )+\left ( y^{2}+6y +9\right )= 11+4+9  (Adding and subtracting 4 and 9 respectively to complete the square)
\left ( x-2 \right )^{2}+\left ( y+3 \right )^{2}= 24

So, the standard form of the equation of the circle is \left ( x-2 \right )^{2}+\left ( y+3 \right )^{2}= 24
Here,
h= 2
k= -3
To calculate the radius of the circle given by the equation \left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2},

we simply take the square root of both sides, i.e.,
\sqrt{\left [\left ( x-h \right )^{2} +\left ( y-k \right )^{2}\right ]}= r,

In this case, we have \left ( x-2 \right )^{2}+\left ( y+3 \right )^{2}= 24,  so the radius r is given by:
r= \sqrt{\left [\left ( x-2 \right ) ^{2}+\left ( y+3 \right )^{2}\right ]}
r= \sqrt{24}  (since the equation holds for any point on the circle)
r= \sqrt{4\times 6}
r= 2\sqrt{6}

Therefore, the radius of the circle is 2\sqrt{6}.

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Pankaj

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