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

1 Answer
Tony B
Jun 26, 2016

#x^2+y^2=10sqrt(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 #(10,10)# as though it is the hypotenuse.

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

Then #r^2=x^2+y^2#

#=>r=sqrt(10^2+10^2) = sqrt(200)#

#=>r=sqrt(2xx10^2)=10sqrt(2)#

So the equation of the circle is:

#x^2+y^2=10sqrt(2)#

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