Question

a) Find the center and the radius of the circle, please show all ofyour work and the equation of the circle in center radius form.x2 + y2 + 6x + 8y - 75 = 0Center Radius Form: (x-h)^2 + (y-k)^2 = r^2Center: Radius:

Answer 1

Let's copy the given equation:

`x^2+y^2+6x+8y-75=0`

To right in the center-radius form, we will have to complete the square for "x" and for "y".

The square binomial is as follows:

`(x+a)^2=x^2+2ax+a^2`

Let's group the "x" terms and the "y" terms:

`(x^2+6x)+(y^2+8y)-75=0`

We can see, for "x", that the quadratic term is already as we need and we have, for the first degree term:

`\begin{gathered} 6=2a \\ a=(6)/(2) \\ a=3 \end{gathered}`

So, if a = 3, we know that the 0th degree term is:

`\begin{gathered} a^2=(3)^2 \\ a^2=9 \end{gathered}`

So, to complete the square, we will add 9 and, to don't change the final value, substract 9:

`\begin{gathered} (x^2+6x+9-9)+(y^2+8y)-75=0 \\ (x^2+6x+9)-9+(y^2+8y)-75=0 \\ (x+3)^2+(y^2+8y)-84=0 \end{gathered}`

Now, we do the same for "y". Let's use "b" this time to don't get confused with "a". The quadratic term is correct and from the 1st degree term we get:

`\begin{gathered} 8=2b \\ b=(8)/(2) \\ b=4 \end{gathered}`

So, we know that the 0th term is:

`\begin{gathered} b^2=4^2 \\ b^2=16 \end{gathered}`

So, we add and substract 16 to comlpete the square:

`\begin{gathered} (x+3)^2+(y^2+8y+16-16)-84=0 \\ (x+3)^2+(y^2+8y+16)-16-84=0 \\ (x+3)^2+(y+4)^2-100=0 \end{gathered}`

Now, we put the third term, which has the radius information, to the other side of the equation and take the square root and the square:

`\begin{gathered} (x+3)^2+(y+4)^2=100 \\ (x+3)^2+(y+4)^2=(\sqrt[]{100})^2 \\ (x+3)^2+(y+4)^2=10^2 \end{gathered}`

Comparing with the center-radius form:

`(x-h)^2+(y-k)^2=r^2`

We get:

`\begin{gathered} h=-3 \\ k=-4 \\ r=10 \end{gathered}`

Since the center is C = (h, k), the answer is:

Center: (-3, -4)

Radius: 10