The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is
Show Hint
To save time on an exam, you can simply plug the given coordinates $(3,7)$ and $(5,5)$ directly into the four options! The correct equation must evaluate to $0 = 0$ for both points.
Step 1: Use the centre on the line.
Let the centre be $(h,k)$. Since it lies on $x - 4y = 1$, we have $h = 4k + 1$.
Step 2: Equal distances to both points.
The centre is the same distance from $(3,7)$ and $(5,5)$. Setting the squared distances equal and using $h = 4k+1$ gives $k = -1$, then $h = -3$.
Step 3: Find the radius and write the circle.
$r^2 = (-3-5)^2 + (-1-5)^2 = 64 + 36 = 100$. So $(x+3)^2 + (y+1)^2 = 100$, which expands to $x^2 + y^2 + 6x + 2y - 90 = 0$.
\[ \boxed{x^2 + y^2 + 6x + 2y - 90 = 0} \]