How do you graph $x^{2} + y^{2} - 8 x - 6 y - 11 = 0$ $x^2 + y^2 - 8x - 6y - 11 = 0$ ?

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How do you graph $x^{2} + y^{2} - 8 x - 6 y - 11 = 0$ $x^2 + y^2 - 8x - 6y - 11 = 0$ ?

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Answer 1 · 2015-09-26T23:03:00

Refer to explanation

Explanation:

The equation $x^{2} + y^{2} - 8 x - 6 y - 11 = 0$ $x^2 + y^2 - 8x - 6y - 11 = 0$ can be rewritten as follows

$(x^{2} - 8 x + 16) + (y^{2} - 6 y + 9) = 11 + 16 + 9 \Rightarrow {(x - 4)}^{2} + {(y - 3)}^{2} = 6^{2}$ $(x^2-8x+16)+(y^2-6y+9)=11+16+9=> (x-4)^2+(y-3)^2=6^2$

which is a circle with center $(4, 3)$ $(4,3)$ and radius $r = 6$ $r=6$ hence

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How do you graph $x^{2} + y^{2} - 8 x - 6 y - 11 = 0$ $x^2 + y^2 - 8x - 6y - 11 = 0$ ?

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Explanation:

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How do you graph $x^{2} + y^{2} - 8 x - 6 y - 11 = 0$ $x^2 + y^2 - 8x - 6y - 11 = 0$ ?