The coordinates of the centre of a circle are $(x - 7, 2x)$. Find the value(s) of $'x'$, if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units.
Show Hint
Always divide by the greatest common divisor (here, 5) as early as possible to make the coefficients of the quadratic equation smaller and easier to factorize.
This reduces chances of arithmetic mistakes during factorization.
Step 1: Set up the same distance equation, but finish it with the quadratic formula instead of factoring.
The centre of the circle is $(x - 7, 2x)$, it passes through $(-9, 11)$, and the radius is $5\sqrt{2}$. Since every point on a circle is exactly one radius away from the centre, the distance from the centre to $(-9, 11)$ must equal $5\sqrt{2}$.
Step 2: Write the distance equation and square both sides.
\[ \sqrt{[(-9) - (x-7)]^2 + [11 - 2x]^2} = 5\sqrt{2} \]
Squaring removes the square root:
\[ [(-9) - (x - 7)]^2 + (11 - 2x)^2 = 50 \]
Step 3: Expand each bracket fully, without shortcuts.
The first bracket:
\[ (-9) - (x - 7) = -9 - x + 7 = -x - 2 \]
\[ (-x-2)^2 = x^2 + 4x + 4 \]
The second bracket:
\[ (11 - 2x)^2 = 121 - 44x + 4x^2 \]
Step 4: Combine and simplify into a standard quadratic.
\[ (x^2 + 4x + 4) + (4x^2 - 44x + 121) = 50 \]
\[ 5x^2 - 40x + 125 = 50 \]
\[ 5x^2 - 40x + 75 = 0 \]
Divide throughout by 5:
\[ x^2 - 8x + 15 = 0 \]
Step 5: Solve using the quadratic formula instead of splitting the middle term.
Here $a = 1$, $b = -8$, $c = 15$.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm \sqrt{4}}{2} = \frac{8 \pm 2}{2} \]
This gives:
\[ x = \frac{8+2}{2} = 5 \qquad \text{or} \qquad x = \frac{8-2}{2} = 3 \]
Step 6: Final answer.
The values of $x$ are 3 and 5.
\[ \boxed{x = 3 \text{ or } x = 5} \]
