Question

What is the center and radius of the circle with equation `x^2 + y^2 - 2x + 14y + 42 = 0`?

Answer 1

the centre is `(1,-7)` and the radius is `sqrt8=2sqrt2`

Explanation:

`x^2+y^2-2x+14y+42=0`

`(x^2-2x)+(y^2+14y)=-42`

Complete the square. If you have an equation `ax^2+bx` and you want to complete the square, all you have to do is to half the coefficient of the term `x` ie `b/2` and then square it ie `(b/2)^2`. Your equation will end up being `ax^2+bx+(b/2)^2`

`(x^2-2x+1)+(y^2+14y+49)=-42+1+49`

`(x-1)^2+(y+7)^2=8`

Since the general form of the circle is given by: `(x-h)^2+(y-k)^2=r^2` where the centre is `(h,k)` and the radius is r, then looking at the above equation, the centre is `(1,-7)` and the radius is `sqrt8=2sqrt2`