Question

Prove whether or not the point (√21,2) lies on a circle centered at the origin and containing the point (5,0).

Answer 1

check the picture below.

so, we know the radius of this circle is 5 then, namely, the distance from (0,0) to (5,0) is 5.

now, if (√(21) , 2) indeed lies on that circle curve, then the distance from (0,0) to (√(21) , 2) will also be the same radius of 5 units.

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{\textit{origin}}{(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})}\qquad (\stackrel{x_2}{√(21)}~,~\stackrel{y_2}{2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ r=\sqrt{(√(21)-0)^2+(2-0)^2}\implies r=\sqrt{(√(21))^2+2^2} \\\\\\ r=√(21+4)\implies r=√(25)\implies r=5~~~~\checkmark`

Answer 2

It does. "Proof" below.

Explanation:

In order for the given points to lie on the circle, they must both be the same distance from the origin. The Pythagorean theorem is used to make a "distance formula" for computing the distance between two points.

For a point (x, y), its distance (d) to the origin will be ...

... d = √(x² +y²)

For the reference point that we know is on the circle, this distance (the circle's radius) is ...

... d = √(5² +0²) = √25 = 5

For the point in question, the distance to the origin is ...

... d = √((√21)² +2²) = √(21 +4) = √25 = 5

Both points have the same distance to the origin, 5 units, so a circle (of radius 5) centered there will contain both points.

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A graph is not proof, but it can confirm the result.