Question

2x² + 2y² + 4x + 16y + 26 = 0 is the equation of a circle with center (h, k) and radius r for:
h =
and
k =
and
r =
question help: video written example

Explanation:

Step1: Divide by 2 to simplify

Divide the entire equation \(2x^{2}+2y^{2}+4x + 16y+26 = 0\) by 2. We get \(x^{2}+y^{2}+2x + 8y+13=0\).

Step2: Group x and y terms

Group the \(x\)-terms and \(y\)-terms: \((x^{2}+2x)+(y^{2}+8y)= - 13\).

Step3: Complete the square for x

For the \(x\)-terms \(x^{2}+2x\), we add \((\frac{2}{2})^{2}=1\) to complete the square. So \(x^{2}+2x + 1=(x + 1)^{2}\).

Step4: Complete the square for y

For the \(y\)-terms \(y^{2}+8y\), we add \((\frac{8}{2})^{2}=16\) to complete the square. So \(y^{2}+8y + 16=(y + 4)^{2}\).

Step5: Add constants to both sides

Add 1 and 16 to the right - hand side as well (since we added them to the left - hand side). We have \((x + 1)^{2}+(y + 4)^{2}=-13 + 1+16\).

Step6: Simplify the right - hand side

Simplify the right - hand side: \(-13 + 1+16=4\). The equation of the circle in standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\) is \((x+1)^{2}+(y + 4)^{2}=2^{2}\). Comparing with the standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\), we get \(h=-1\), \(k = - 4\) and \(r = 2\).

Answer:

\(h=-1\), \(k=-4\), \(r = 2\)