What is the general form of the equation of a circle with a center at (7, 0) and a radius of 10?

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Answer 1 · 2015-11-12T23:26:37

$x^{2} - 14 x + y^{2} - 51 = 0$ $x^2 - 14x + y^2 - 51 = 0$

First, let's write the equation in standard form.

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

$\Rightarrow {(x - 7)}^{2} + {(y - 0)}^{2} = 10^{2}$ $=> (x - 7)^2 + (y - 0)^2 = 10^2$

$\Rightarrow {(x - 7)}^{2} + y^{2} = 10^{2}$ $=> (x - 7)^2 + y^2 = 10^2$

Then, we expand the equation.

$\Rightarrow (x^{2} - 14 x + 49) + y^{2} = 100$ $=> (x^2 - 14x + 49) + y^2 = 100$

Finally, let's put all the terms in one side and simplify

$\Rightarrow x^{2} - 14 x + 49 + y^{2} - 100 = 0$ $=> x^2 -14x + 49 + y^2 - 100 = 0$

$\Rightarrow x^{2} - 14 x + y^{2} - 51 = 0$ $=> x^2 - 14x + y^2 - 51 = 0$