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 square the distance formula first to avoid carrying the square root symbol through all your calculations.
This makes the algebraic expansion much cleaner and reduces clerical errors.
Step 1: Set up the same distance condition.
The distance from the centre $C(x-7, 2x)$ to the point $P(-9, 11)$ on the circle must equal the radius $5\sqrt{2}$. Squaring both sides of the distance formula avoids dealing with the square root:
\[ (x - 7 - (-9))^2 + (2x - 11)^2 = (5\sqrt{2})^2 \]
Step 2: Simplify each bracket.
\[ (x + 2)^2 + (2x - 11)^2 = 50 \]
\[ (x^2 + 4x + 4) + (4x^2 - 44x + 121) = 50 \]
Step 3: Collect like terms into one quadratic equation.
\[ 5x^2 - 40x + 125 = 50 \]
\[ 5x^2 - 40x + 75 = 0 \]
Dividing throughout by 5:
\[ x^2 - 8x + 15 = 0 \]
Step 4: Solve using the quadratic formula instead of factoring.
Here $a = 1$, $b = -8$, $c = 15$.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm \sqrt{4}}{2} = \frac{8 \pm 2}{2} \]
Step 5: Work out both values.
\[ x = \frac{8 + 2}{2} = 5 \qquad \text{or} \qquad x = \frac{8 - 2}{2} = 3 \]
Final Answer:
The possible values of $x$ are 3 and 5.
\[ \boxed{x = 3 \text{ or } x = 5} \]
