Question

question
find the center and radius of the circle represented by the equation below.
$x^{2}+y^{2}+8x - 20y + 35 = 0$
answer attempt 1 out of 2
center: (□,□)
radius: □

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together: \(x^{2}+8x + y^{2}-20y=-35\)

Step2: Complete the square for x

For the \(x\)-terms: \(x^{2}+8x\), take half of 8 (which is 4), square it (\(4^{2} = 16\)). Add 16 to both sides.
For the \(y\)-terms: \(y^{2}-20y\), take half of -20 (which is -10), square it (\((-10)^{2}=100\)). Add 100 to both sides.

So we have:
\(x^{2}+8x + 16+y^{2}-20y + 100=-35 + 16+100\)

Step3: Rewrite as perfect squares

Rewrite the left side as perfect squares: \((x + 4)^{2}+(y - 10)^{2}=81\)

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Comparing \((x + 4)^{2}+(y - 10)^{2}=81\) with the standard form, we get \(h=-4\), \(k = 10\), and \(r^{2}=81\), so \(r = 9\) (since radius is positive).

Answer:

Center: \((-4, 10)\)
Radius: \(9\)