# Question de290

Jan 11, 2017

#### Answer:

(x-4)^2 +(y+5)^2 =36= 6^2

#### Explanation:

Rewrite as ${x}^{2} - 8 x + {y}^{2} + 10 y + 5 = 0$

${x}^{2} - 8 x + 16 - 16 + {y}^{2} + 10 y + 25 - 25 + 5$

${\left(x - 4\right)}^{2} + {\left(y + 5\right)}^{2} - 16 - 25 + 5 = 0$

${\left(x - 4\right)}^{2} + {\left(y + 5\right)}^{2} = 36 = {6}^{2}$

Jan 11, 2017

#### Answer:

You complete the squares, using the patterns:
${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2} \text{ [1]}$
and
${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2} \text{ [2]}$

#### Explanation:

The standard form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [3]}$

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

Given: ${x}^{2} + {y}^{2} - 8 x + 10 y + 5 = 0 \text{ [4]}$

Move the x terms together, the y terms together, and the constant to the right:

${x}^{2} - 8 x + {y}^{2} + 10 y = - 5 \text{ [4]}$

We want to make the first 3 terms in equation [4] look like the right side of equation [1] so we insert an ${h}^{2}$ as the third term but, to keep the equation balanced, we must add ${h}^{2}$ on the right:

${x}^{2} - 8 x + {h}^{2} + {y}^{2} + 10 y = {h}^{2} - 5 \text{ [5]}$

The first 3 terms of equation [5] look like the right side of equation [1].

We can match the $- 2 h x$ in equation [1] with the $- 8 x$ in equation [5] and write the equation:

$- 2 h x = - 8 x$

Find the value of h by dividing both sides of the equation by -2x:

$h = 4$

This means that, in equation [5], we can replace the terms ${x}^{2} - 8 x + {h}^{2}$ with ${\left(x - 4\right)}^{2}$ and replace the ${h}^{2}$ on the right with 16:

${\left(x - 4\right)}^{2} + {y}^{2} + 10 y = 16 - 5 \text{ [6]}$

We want to make the y terms in equation [6] look like the right side of equation [2} so we add a ${k}^{2}$ on the left but, to keep the equation balanced, we must add a ${k}^{2}$ to the right side:

${\left(x - 4\right)}^{2} + {y}^{2} + 10 y + {k}^{2} = {k}^{2} + 16 - 5 \text{ [7]}$

The y terms on the left of equation [7] look like the right side of equation [2].

Match the $- 2 k y$ in equation [2] with the $+ 10 y$ in equation [7] and write the equation:

$- 2 k y = + 10 y$

Find the value of k by dividing both sides by -2y:

$k = - 5$

This means that, in equation [7], we can replace ${y}^{2} + 10 y + {k}^{2}$ with ${\left(y - - 5\right)}^{2}$ and the k^2 on the right with 25:

${\left(x - 4\right)}^{2} + {\left(y - - 5\right)}^{2} = 25 + 16 - 5 \text{ [8]}$

Simplify the constants on the right:

${\left(x - 4\right)}^{2} + {\left(y - - 5\right)}^{2} = 36 \text{ [9]}$

Write the constant as a square:

${\left(x - 4\right)}^{2} + {\left(y - - 5\right)}^{2} = {6}^{2} \text{ [10]}$

This is a circle with a radius of 6 and a center at $\left(4 , - 5\right)$