Question

1. write the equation of the circle below in standard form.\\(x^2 + y^2 - 14x + 4y - 11 = 0\\)\\((quad)^2 + (quad)^2 = quad\\)\\(\text{center} = (quad, quad)\\)\\(\text{radius} = quad\\)

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together: \(x^{2}-14x + y^{2}+4y=11\)

Step2: Complete the square for x

For the \(x\)-terms: \(x^{2}-14x\). Take half of \(-14\) which is \(-7\), square it: \((-7)^{2} = 49\). Add 49 to both sides.

Step3: Complete the square for y

For the \(y\)-terms: \(y^{2}+4y\). Take half of \(4\) which is \(2\), square it: \(2^{2}=4\). Add 4 to both sides.

Step4: Rewrite as perfect squares

After adding, we get: \((x - 7)^{2}+(y + 2)^{2}=11 + 49+4\)

Step5: Calculate the right-hand side

Calculate \(11 + 49+4=64\). So the standard form is \((x - 7)^{2}+(y + 2)^{2}=64\)

Step6: Identify center and radius

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. So center is \((7,-2)\) and radius \(r=\sqrt{64} = 8\)

Answer:

Standard Form: \((x - 7)^{2}+(y + 2)^{2}=64\)
Center \(=(7,-2)\)
Radius \(= 8\)