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Prove or disprove that the point (51−−√, 7) lies on the circle centered at the origin and containing the point (6, 8).

User Anlis
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1 Answer

2 votes

Answer:

Yes, it lies on the circle.

Explanation:

The actual question is:

"Prove or disprove that the point (√51, 7) lies on the circle centered at the origin and containing the point (6, 8)."

The equation of a circle of radius R and centered at the point (a, b) is:

(x - a)^2 + (y - b)^2 = R^2

In this case we have a circle centered at the origin, then: (a, b) = (0, 0)

And the equation becomes:

(x - 0)^2 + (y - 0)^2 = R^2

x^2 + y^2 = R^2

We also know that this circle contains the point (6, 8), then we can find the radius of our circle if we replace x by 6 and y by 8.

6^2 + 8^2 = R^2

100 = R^2

√100 = 10 = R

Then the circle equation is:

x^2 + y^2 = 10^2

Now let's see if the point (√51, 7) is a solution of the above equation:

(√51)^2 + ( 7)^2 = 10^2 = 100

51 + 49 = 100

100 = 100

This is true, so we can conclude that (√51, 7) is a solution of the circle equation, so the point lies on the defined circle.

User Kevin Mei
by
8.0k points
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