How do you graph $X^{2} + Y^{2} - 16 X + 4 Y + 52 = 0$ $X^2+Y^2-16X+4Y+52=0$ ?

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Answer 1 · 2017-01-14T00:26:32

This is a circle of radius $4$ $4$ with centre $(8, - 2)$ $(8, -2)$

Complete the square for both $x$ $x$ and $y$ $y$ in order to get this in the form of the standard equation of a circle:

Given:

$x^{2} + y^{2} - 16 x + 4 y + 52 = 0$ $x^2+y^2-16x+4y+52=0$

Reorganise as:

$x^{2} - 16 x + 64 + y^{2} + 4 y + 4 - 16 = 0$ $color(blue)(x^2-16x+64)+color(green)(y^2+4y+4)-16=0$

That is:

$x^{2} - 2 (8 x) + 8^{2} + y^{2} + 2 (2 y) + 2^{2} - 4^{2} = 0$ $color(blue)(x^2-2(8x)+8^2)+color(green)(y^2+2(2y)+2^2)-4^2=0$

Hence:

${(x - 8)}^{2} + {(y + 2)}^{2} = 4^{2}$ $color(blue)((x-8)^2)+color(green)((y+2)^2) = 4^2$

which is (more or less) in the form:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $color(blue)((x-h)^2)+color(green)((y-k)^2) = r^2$

with $(h, k) = (8, - 2)$ $(h, k) = (8, -2)$ being the centre of the circle and $r = 4$ $r = 4$ being the radius.

graph{(x^2+y^2-16x+4y+52)((x-8)^2+(y+2)^2-0.038)=0 [-4.08, 15.92, -6.84, 3.16]}

How do you graph X2+Y2−16X+4Y+52=0X^2+Y^2-16X+4Y+52=0?