Question

question help: video written example
2x² + 2y² + 16x + 20y - 16 = 0 is the equation of a circle with center (h, k) and radius r for
h=
and k=
and r=

Explanation:

Step1: Convert the given equation to the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\).
First, divide the given equation \(2x^{2}+2y^{2}+16x + 20y-16 = 0\) by \(2\):
\(x^{2}+y^{2}+8x + 10y-8=0\).
Then, complete the square for \(x\) and \(y\) terms.
For the \(x\) - terms: \(x^{2}+8x=(x + 4)^{2}-16\).
For the \(y\) - terms: \(y^{2}+10y=(y + 5)^{2}-25\).
Substitute these into the equation:
\((x + 4)^{2}-16+(y + 5)^{2}-25-8=0\).
Simplify to get \((x + 4)^{2}+(y + 5)^{2}=49\).

Step2: Identify \(h\), \(k\) and \(r\)

Comparing \((x + 4)^{2}+(y + 5)^{2}=49\) with \((x - h)^2+(y - k)^2=r^2\).
We have \(h=-4\), \(k = - 5\), and \(r = 7\) (since \(r^{2}=49\) and \(r>0\)).

Answer:

\(h=-4\), \(k=-5\), \(r = 7\)