Find the equation of the circle with centre at (2,6) which passes through the point (3,7)

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Given the center $(h, k) = (2, 6)$, the equation becomes:

$$(x - 2)^2 + (y - 6)^2 = r^2$$

The circle passes through the point $(-3, 7)$. We can substitute these coordinates into the equation from Step 1 to find $r^2$.

$$(-3 - 2)^2 + (7 - 6)^2 = r^2$$ $$(-5)^2 + (1)^2 = r^2$$ $$25 + 1 = r^2$$ $$r^2 = 26$$

Substitute the value of $r^2$ back into the equation from Step 1.

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