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

May 1, 2018

The equation of the circle in standard form is ${\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = 25$
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. ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
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.
${x}^{2} - 8 x + {y}^{2} - 4 y - 5 = 0$

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.

$\implies {x}^{2} - 8 x + 16 + {y}^{2} - 4 y + 4 - 5 - 16 - 4 = 0$

Note that there are 2 polynomials of the form ${a}^{2} - 2 a b + {b}^{2.}$
Write them in the form of ${\left(a - b\right)}^{2}$.
$\implies {\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} - 25 = 0 \implies {\left(x - 4\right)}^{2} + {\left(y - 2\right)}^{2} = 25$

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