Question

find the center and radius of the circle represented by the equation below.

$x^{2}+y^{2}-4x + 16y + 19 = 0$

answer attempt 1 out of 10

center: $()$

radius: $()$

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Explanation:

Step1: Rearrange and group x and y terms

$$x^{2}-4x + y^{2}+16y=-19$$

Step2: Complete the square for x - terms

For \(x^{2}-4x\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2ab = 4x\Rightarrow b = 2\). Then \(x^{2}-4x=(x - 2)^{2}-4\)

Step3: Complete the square for y - terms

For \(y^{2}+16y\), using \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = y\), \(2ab=16y\Rightarrow b = 8\). Then \(y^{2}+16y=(y + 8)^{2}-64\)

Step4: Substitute back into the equation

\((x - 2)^{2}-4+(y + 8)^{2}-64=-19\)
\((x - 2)^{2}+(y + 8)^{2}-68=-19\)
\((x - 2)^{2}+(y + 8)^{2}=49\)

Answer:

Center: \((2,-8)\)
Radius: \(7\)