Question

for the equation $2x^2 + 2y^2 - 16x + 4y - 16 = 0$, do the following.
(a) find the center $(h,k)$ and radius $r$ of the circle.
(b) graph the circle.
(c) find the intercepts, if any.
(a) the center is $(4, -1)$.
(simplify your answer. type an ordered pair.)
the radius is $r = $
(simplify your answer.)

Explanation:

Step1: Divide equation by 2

Given equation \(2x^{2}+2y^{2}-16x + 4y-16 = 0\), divide each term by 2:
\(x^{2}+y^{2}-8x + 2y-8 = 0\)

Step2: Group x and y terms

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

Step3: Complete the square for x

For \(x^{2}-8x\), take half of \(-8\) (\(-4\)), square it (\(16\)):
\(x^{2}-8x + 16=(x - 4)^{2}\)

Step4: Complete the square for y

For \(y^{2}+2y\), take half of \(2\) (\(1\)), square it (\(1\)):
\(y^{2}+2y + 1=(y + 1)^{2}\)

Step5: Add squares to both sides

Add \(16\) (from x) and \(1\) (from y) to both sides:
\((x - 4)^{2}+(y + 1)^{2}=8 + 16+1\)
Simplify right side: \(8 + 16+1 = 25\)

Step6: Identify radius

The standard circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\). Here, \(r^{2}=25\), so \(r=\sqrt{25}=5\).

Answer:

\(5\)