# How do you find the center and radius of the circle  x^2-2x+y^2-8y+1=0?

Dec 19, 2016

Radius: $4$
Centre: $\left(- 1 , 4\right)$

#### Explanation:

You should convert to the form ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$. Complete the square twice to achieve this.

$1 \left({x}^{2} - 2 x + n\right) + 1 \left({y}^{2} - 8 y + m\right) = - 1$

Our goal here is to make the expressions in parentheses into perfect squares. We can use the formula $n = {\left(\frac{b}{2}\right)}^{2}$ and $m = {\left(\frac{b}{2}\right)}^{2}$, where $b$ is the x-coefficent term.

So,

$n = {\left(- \frac{2}{2}\right)}^{2} = 1$

$m = {\left(- \frac{8}{2}\right)}^{2} = 16$

We need to add and subtract this number inside the parentheses to keep the equation equivalent.

$1 \left({x}^{2} - 2 x + 1 - 1\right) + 1 \left({y}^{2} - 8 x + 16 - 16\right) = - 1$

$1 \left({x}^{2} - 2 x + 1\right) - 1 + 1 \left({y}^{2} - 8 x + 16\right) - 16 = - 1$

${\left(x - 1\right)}^{2} + {\left(y - 4\right)}^{2} = - 1 + 17$

${\left(x + 1\right)}^{2} + {\left(y - 4\right)}^{2} = 16$

The radius of a circle in the form ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ is $r$, so in this case $4$. The centre is at $\left(a , b\right)$, in this case $\left(- 1 , 4\right)$.

Hopefully this helps!

Dec 19, 2016

You complete the squares by adding ${h}^{2} \mathmr{and} {k}^{2}$ to both sides of the equation. Please see the explanation.

#### Explanation:

The standard Cartesian form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ }$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and $r$ is the radius.

Equation  is the same as equation  but with the squares expanded and equation  is the given equation with some spaces added for missing terms:

${x}^{2} - 2 h x + {h}^{2} + {y}^{2} - 2 k y + {k}^{2} = {r}^{2} \text{ }$
${x}^{2} - 2 x \text{ " + y^2 - 8y" "+ 1= 0" }$

Let's try to make equations  and  match by adding ${h}^{2} \mathmr{and} {k}^{2}$ to both sides of equation  and label it equation :

${x}^{2} - 2 h x + {h}^{2} + {y}^{2} - 2 k y + {k}^{2} = {r}^{2} \text{ }$
${x}^{2} - 2 x + {h}^{2} + {y}^{2} - 8 y + {k}^{2} + 1 = {h}^{2} + {k}^{2} \text{ }$

Subtract 1 from both sides of equation  and label it equation :

${x}^{2} - 2 h x + {h}^{2} + {y}^{2} - 2 k y + {k}^{2} = {r}^{2} \text{ }$
${x}^{2} - 2 x + {h}^{2} + {y}^{2} - 8 y + {k}^{2} = {h}^{2} + {k}^{2} - 1 \text{ }$

Now that equation  and equation  both have 6 terms on the left side, we can see that we can find the value of h by equating the second term in equation  with the same term in equation :

$- 2 h x = - 2 x$

$h = 1$

Substitute 1 for h into equation  and number it equation :

${x}^{2} - 2 h x + {h}^{2} + {y}^{2} - 2 k y + {k}^{2} = {r}^{2} \text{ }$
${x}^{2} - 2 x + {1}^{2} + {y}^{2} - 8 y + {k}^{2} = {1}^{2} + {k}^{2} - 1 \text{ }$

We can find the value of k by equating the fifth term in equation  with the same term in equation :

$- 2 k y = - 8 y$

$k = 4$

Substitute 4 for k into equation  and number it equation :

${x}^{2} - 2 h x + {h}^{2} + {y}^{2} - 2 k y + {k}^{2} = {r}^{2} \text{ }$
${x}^{2} - 2 x + {1}^{2} + {y}^{2} - 8 y + {4}^{2} = {1}^{2} + {4}^{2} - 1 \text{ }$

Because we used the patterns for ${\left(x - h\right)}^{2} \mathmr{and} {\left(y - k\right)}^{2}$, respectively, we know that the first three terms become ${\left(x - 1\right)}^{2}$ and the next three terms become ${\left(y - 4\right)}^{2}$:

${\left(x - 1\right)}^{2} + {\left(y - 4\right)}^{2} = {1}^{2} + {4}^{2} - 1 \text{ }$

Simplify the right side:

${\left(x - 1\right)}^{2} + {\left(y - 4\right)}^{2} = {4}^{2} \text{ }$
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ }$

Comparing equation  with equation , we can see that the center is $\left(1 , 4\right)$ and the radius is 4.