How do you find the equation for the circle centered at (4, 4); passing through (7, 14)?

Jul 31, 2016

${\left(x - 4\right)}^{2} + {\left(y - 4\right)}^{2} = 109$

or

${x}^{2} + {y}^{2} - 8 x - 8 y - 77 = 0$

Explanation:

The general equation of a circle with centre $\left(h , k\right)$ and radius $r$ may be written:

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

In our example, we know $\left(h , k\right) = \left(4 , 4\right)$, so we just need to determine ${r}^{2}$.

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

Since the circle passes through $\left(7 , 14\right)$, that pair of values for $x$ and $y$ must satisfy the equation and we have:

${r}^{2} = {\left(7 - 4\right)}^{2} + {\left(14 - 4\right)}^{2} = {3}^{2} + {10}^{2} = 109$

So the equation of the circle may be written:

${\left(x - 4\right)}^{2} + {\left(y - 4\right)}^{2} = 109$

We can reexpress this in standard polynomial form by expanding and rearranging:

$\textcolor{b l u e}{{x}^{2} - 8 x + 16} + \textcolor{g r e e n}{{y}^{2} - 8 y + 16} = 109$

So:

${x}^{2} + {y}^{2} - 8 x - 8 y - 77 = 0$