Answer:
Yes, it lies on the circle.
Step-by-step 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.
