Question

How do you find the radius of a circle with the equation `x^2 - 8x + y^2 - 4y – 5 = 0`?

Answer 1

The equation of the circle in standard form is `(x-4)^2 + (y-2)^2 = 25`
25 is the square of the radius. So the radius must be 5 units. Also, the centre of the circle is (4, 2)

Explanation:

To calculate the radius/centre, we must first convert the equation to standard form. `(x - h)^2 + (y - k)^2 = r^2`
where (h, k) is the centre and r is the radius of the circle.

The procedure to do this would be to complete the squares for x and y, and transpose the constants to the other side.
`x^2 - 8x + y^2 - 4y - 5 = 0`

To complete the squares, take the coefficent of the term with degree one, divide it by 2 and then square it. Now add this number and subtract this number. Here, the coefficient of the terms with degree 1 for x and y are (-8) and (-4) respectively. Thus we must add and subtract 16 to complete the square of x as well as add and subtract 4 to complete the square of y.

`implies x^2 - 8x +16 + y^2 - 4y + 4 - 5 -16 -4 = 0`

Note that there are 2 polynomials of the form `a^2 - 2ab + b^2.`
Write them in the form of `(a - b)^2`.
`implies (x - 4)^2 + (y - 2)^2 - 25= 0 implies (x -4)^2 + (y - 2)^2 = 25`

This is of the standard form. So 25 must be the square of the radius. This means that the radius is 5 units.