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

Here's how to find the equation of the circle:

Step 1: Write the general equation of a circle with the given center. 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: $(x - 2)^2 + (y - 6)^2 = r^2$

Step 2: Calculate the radius squared ($r^2$) using the point the circle passes through. The circle passes through the point $(-3, 7)$. We can substitute these coordinates into the equation from Step 1 to find $r^2$. $(-3 - 2)^2 + (7 - 6)^2 = r^2$ $(-5)^2 + (1)^2 = r^2$ $25 + 1 = r^2$ $r^2 = 26$

Step 3: Write the final equation of the circle. Substitute the value of $r^2$ back into the equation from Step 1. $(x - 2)^2 + (y - 6)^2 = 26$

The equation of the circle is $\boxed{(x - 2)^2 + (y - 6)^2 = 26}$.

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