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What is the centre and radius of the circle with the equation x^{2}+y^{2}+8x+4y+19= 0?

Option: 1

Centre: \left ( -4,-2 \right ) , Radius: 1


Option: 2

Centre: \left ( 4,2 \right ) , Radius: \sqrt{2}


Option: 3

Centre:\left ( -4,-2 \right ) , Radius: \sqrt{3}


Option: 4

Centre: \left ( 4,2 \right ), Radius: \sqrt{3}


Answers (1)

best_answer

To find the centre and radius of the circle with the equation x^{2}+y^{2}+8x+4y+19= 0.
we need to convert it into the standard form of the equation of a circle.

x^{2}+y^{2}+8x+4y+19= 0
x^{2}+y^{2}+8x+4y= -19

Completing the square for x and y, we get:

\left ( x+4 \right )^{2}-16+\left ( y+2 \right )^{2}-4= -19
Simplifying the equation, we get
\left ( x+4 \right )^{2}+\left ( y+2 \right )^{2}= 1

So the centre of the circle is (-4,-2) that means and  h= -4 and k= -2
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}= r^{2} \right ]}= r,

In this case, we have \left ( x+4 \right )^{2}+\left ( y+2 \right )^{2}= 1, so the radius r is given by

r= \sqrt{\left ( x+4 \right )^{2}+\left ( y+2 \right )^{2}}
r= \sqrt{1}  (since the equation holds for any point on the circle)
r= 1

Therefore, the radius of the circle is 1.

Posted by

vishal kumar

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