Circle $C_1$ has equation $x^2 + y^2 - 6x - 2y - 26 = 0$ - Scottish Highers Maths - Question 3 - 2019

To find the radius of circle $C_1$, we first rewrite its equation in standard form. Starting with:

$x^2 + y^2 - 6x - 2y - 26 = 0$

we rearrange it to:

$x^2 - 6x + y^2 - 2y = 26$

Next, we complete the square for the $x$ and $y$ terms. For $x^2 - 6x$, we take half of -6, which is -3, square it to get 9:

$x^2 - 6x = (x - 3)^2 - 9$

For $y^2 - 2y$, we take half of -2, which is -1, square it to get 1:

$y^2 - 2y = (y - 1)^2 - 1$

Now substituting these back, we have:

$(x - 3)^2 - 9 + (y - 1)^2 - 1 = 26$

This simplifies to:

$(x - 3)^2 + (y - 1)^2 = 36$

Thus, the center of circle $C_1$ is $(3, 1)$ and its radius, $r$, is:

$r = ext{sqrt}(36) = 6$