Question 218172
The upper half of a circle would be a function. So we sill start by writing the equation in the form of a function (i.e. we will solve for y):
{{{ x^2+y^2=10 }}}
Subtract {{{x^2}}} from both sides:
{{{y^2 = 10 - x^2}}}
Square root of each side:
{{{y = 0 +- sqrt(10-x^2)}}}
(Sorry about the extra zero. Algebra.com's formula software can't handle +- without something in front of it.)
This is the equation of the entire circle. The upper half would be:
{{{y = sqrt(10-x^2)}}} (and the lower half {{{y = - sqrt(10-x^2)}}}<br>
As for the domain, we are looking for the set of all possible/allowable values for x. There are a couple of ways to find the domain of {{{y = sqrt(10-x^2)}}}:<ol><li>Graphically: Graph the circle (or half-circle) and determine what values x can be from the graph:
{{{drawing(300, 300, -4, 4, -4, 4, grid(1), circle(0, 0, sqrt(10)))}}} From this we should be able to tell that x must be equal to or between -radius and +radius. And what is the radius of the circle? From the equation of the circle we should know that the radius is {{{sqrt(10)}}}</li><li>Algebraically: With the square root, we must ensure that the radicand (the expression inside the square root) is either positive or zero. (IOW we cannot allow the radicand to become negative because there are no Real numbers whose squares are negative.)
So the domain is the set of all numbers which fit:
{{{10 - x^2 >= 0}}}
If you cannot "see" the domain from this then we have to solve this quadratic inequality. Solving quadratic inequalities is not easy. Perhaps the "easy" way is the factor {{{10-x^2}}} using the difference of squares pattern creatively:
{{{10-x^2 = (sqrt(10))^2 - x^2 = (sqrt(10) + x)(sqrt(10) - x)}}}. This gives us:
{{{(sqrt(10) + x)(sqrt(10) - x) >= 0}}}
This says that the product on the left is positive or zero. When are products positive or zero? Answer: When both factors are positive or both factors are negative or when one of the factors is zero. With these particular factors, it is impossible for them both to be negative because:<ul><li>For {{{sqrt(10) + x}}} to be negative, x would have to be less than {{{-sqrt(10)}}}</li><li>If x is less than {{{-sqrt(10)}}} then the other factor, {{{sqrt(10) - x}}}, would have to positive. (Think about subtracting a negative!)</li></ul>So both factors must be positive or one of them must be zero:
{{{sqrt(10) + x >= 0}}} and {{{sqrt(10) - x >= 0}}}
(The "or equal to" covers the possibility of zero factors.)
Solving these for x we get:
{{{x >= -sqrt(10)}}} and {{{x <= sqrt(10)}}}</li></ol>
Either way, the domain, in interval notation, is [{{{-sqrt(10)}}}, {{{sqrt(10)}}}]