Question

What are the center and the radius of the circle 4x2 + 4y2 - 8x + 16y -4 = 0?

Answer 1

Answer:

Center of circle: (1, -2)

Radius: √6

Step-by-step explanation:

First, we need to isolate the constant:

$4x^2 + 4y^2 - 8x + 16y = 4$

Then, we can divide the whole equation by 4:

$x^2+y^2-2x+4y=1$

Then, we can complete the square(reminder: (a+b)^2 = a^2 + 2ab + b^2)

$(x^2-2x)+(y^2+4y)=1\\(x^2-2x+1)+(y^2+4y)=1+1\\(x^2-2x+1)+(y^2+4y+4)=1+1+4\\(x-1)^2+(y+2)^2=6$

Now that we got the standard equation of the circle, we can find the center and radius by using the equation $(x-h)^2+(y-k)^2=r^2$, where (h, k) is the coordinate of the center, and r is the radius.

Therefore, the center of the circle is (1, -2), and the radius of the circle is √6.

Answer 2

The center of the circle is (1,-2) and the radius is $\sqrt{6}$.

Important information:

• The given equation of a circle is $4x^2+4y^2-8x+16y-4=0$.

Circle:

The standard equation of the circle is:

$(x-h)^2+(y-k)^2=r^2$            ...(i)

Where, $(h,k)$ is center and $r$ is the radius.

The given equation can be written as:

$4(x^2+y^2-2x+4y-1)=0$

$(x^2-2x)+(y^2+4y)=1$

Add square of half of coefficients of $x$ and $y$.

$\left(x^2-2x+1^2\right)+(y^2+4y+2^2)=1+1^2+2^2$

$(x-1)^2+(y+2)^2=1+1+4$

$(x-1)^2+(y+2)^2=6$

$(x-1)^2+(y+2)^2=(\sqrt{6})^2$           ...(ii)

On comparing (i) and (ii), we get

$h=1$

$k=-2$

$r=\sqrt{6}$

Therefore, the center of the circle is (1,-2) and the radius is $\sqrt{6}$.

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