Question #de290

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Question #de290

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Answer 1 · 2017-01-11T03:48:00

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

Explanation:

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

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

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

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

Answer 2 · 2017-01-11T04:03:06

You complete the squares, using the patterns:
${(x - h)}^{2} = x^{2} - 2 h x + h^{2} [1]$ $(x - h)^2 = x^2 - 2hx + h^2" [1]"$
and
${(y - k)}^{2} = y^{2} - 2 k y + k^{2} [2]$ $(y - k)^2 = y^2 - 2ky + k^2" [2]"$

Explanation:

The standard form for the equation of a circle is:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2} [3]$ $(x - h)^2 + (y - k)^2 = r^2" [3]"$

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

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

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

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

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

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

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

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

$- 2 h x = - 8 x$ $-2hx = -8x$

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

$h = 4$ $h = 4$

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

${(x - 4)}^{2} + y^{2} + 10 y = 16 - 5 [6]$ $(x - 4)^2 + y^2 + 10y = 16 -5" [6]"$

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

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

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

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

$- 2 k y = + 10 y$ $-2ky = +10y$

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

$k = - 5$ $k = -5$

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

${(x - 4)}^{2} + {(y - - 5)}^{2} = 25 + 16 - 5 [8]$ $(x - 4)^2 + (y - -5)^2 = 25 + 16 -5" [8]"$

Simplify the constants on the right:

${(x - 4)}^{2} + {(y - - 5)}^{2} = 36 [9]$ $(x - 4)^2 + (y - -5)^2 = 36" [9]"$

Write the constant as a square:

${(x - 4)}^{2} + {(y - - 5)}^{2} = 6^{2} [10]$ $(x - 4)^2 + (y - -5)^2 = 6^2" [10]"$

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

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Question #de290

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