# How do you find the equation of a circle center at the origin; passes through (10, 10)?

Jun 26, 2016

${x}^{2} + {y}^{2} = 10 \sqrt{2}$

#### Explanation:

You can use Pythagoras for this by treating the x-axis and y-axis as if they were sides of a triangle and a line from the origin to the point $\left(10 , 10\right)$ as though it is the hypotenuse.

Let the hypotenuse be the radius (r) of the circle

Then ${r}^{2} = {x}^{2} + {y}^{2}$

$\implies r = \sqrt{{10}^{2} + {10}^{2}} = \sqrt{200}$

$\implies r = \sqrt{2 \times {10}^{2}} = 10 \sqrt{2}$

So the equation of the circle is:

${x}^{2} + {y}^{2} = 10 \sqrt{2}$
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