Question

What is the equation of the circle with center (0, 0) that passes through the point (5, -4)?

Answer 1

The standard form of a circle is $(x-h) ^{2} +(y-k) ^{2} = r^{2}$ where h and k are the center and x and y are the coordinates you're given.  We need to solve for the radius to finish this off correctly.  Filling in, we have $(5-0) ^{2} +(-4-0) ^{2} = r^{2}$ and $25+16= r^{2}$, giving us that $r^{2}=41$.  Therefore, our equation is $x^{2} + y^{2} =41$

Answer 2

Answer "and" Explanation:

The standard form equation for a circle is:

(x-h)² + (y-k)² = r²

where (h,k) is the center and r is the radius

given that this circle is centered about the origin (0,0), we know that the left half of the equation will be (x-0)² + (y-0)², or just x² + y²

the next step is to find the radius, we do this by using the distance formula to find the distance between a given point and the center of the circle.

the distance formula is:

D = √((x_2-x_1)²+(y_2-y_1)²)

Where given points A and B are defined as (x_1, y_1) and (x_2, y_2) respectively our two given points are the center (0,0) and (5, -4)

so we set our equation up like this:

D = √((5–0)² + (-4–0)²)

And simplify:

D = √((5)² + (-4)²)

D = √(25 + 16)

D = √41

So now we know that √41 is our radius (r-value) and now we can solve the right half of our equation

we need to square √41 and obviously, a square root squared is just the number inside of the radical

(√41)² = 41

so, now we know that our equation for a circle with the given values is:

x² + y² = 41