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

Answer 1

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

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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Answer 2

Greyson Brown

The equation of a circle with center at the origin and passing through the point (10, 10) is:

(x^2 + y^2 = r^2)

Where (r) is the radius of the circle. To find the radius, you can use the distance formula:

(r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2})

Substitute the coordinates of the point (10, 10) into the distance formula along with the origin (0, 0) to find the radius.

(r = \sqrt{(10 - 0)^2 + (10 - 0)^2})

(r = \sqrt{100 + 100})

(r = \sqrt{200})

Therefore, the equation of the circle is:

(x^2 + y^2 = 200)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.