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If the length of a chord of a circle with equation $x^2+y^2=100$ is $16$ units, how far is the chord from the centre?

My Attempt;

$$x^2+y^2=100$$ $$x^2+y^2=10^2$$

So, centre of circle $=(0,0)$.

How do I move further? Please help.

Thanks.

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Without loss of generality, suppose the chord is vertical, and to the right of the origin. Then its $x$ coordinate is the distance from the center. You also know that the endpoints of the chord, $(x,\pm8)$ are on the circle -- this gives you a quadratic equation in $x$ that you can solve: $$ x^2+8^2=100 $$

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    @@Henning, What is the meaning of 'Without loss of generality'? – 2017-02-04
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    @NeWtoN: If the cord is _not_ vertical, then suppose you rotate the entire figure such that is _is_ vertical. This produces another chord also of length 16, and since rotation is a rigid movement, the new chord has the same distance from the center as the old one had. So it is sufficient to find the center distance of the _new_ chord. – 2017-02-04
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    @@Henning, Where does the co ordinate $(x,\pm 8)$ come from? – 2017-02-04
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    @NeWtoN: When the chord is vertical it is perpendicular to the x-axis, which is a diameter in the circle. A diameter that is perpendicular to a chord cuts that chord in half. So half of the 16 units of chord are above the x-axis and the other half are below it. – 2017-02-04
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    Without resorting to WLOGs and the like, one could just note that the perpendicular from the center of the radius to the chord hits the chord in the middle, so you if you also draw the radius to one of the end points of the chord, you get a right triangle with hypotenuse $10$ and one leg $8$. – 2017-02-04
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    @Arthur: Yes, that way of looking at it may be easier (or not, depending on whether one is more comfortable with Euclidean or analytic geometry). – 2017-02-04