Question

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.

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.

Answer 1

Step 1: Understanding the Question:
The centre of a circle is represented by the coordinate point $C(x - 7, 2x)$.
We are told the circle passes through the point $P(-9, 11)$ and its radius is $R = 5\sqrt{2}$ units.
We need to find the possible values of the variable $x$.

Step 2: Key Formula or Approach:
The distance from the centre of a circle to any point on its circumference is equal to the radius.
Using the Distance Formula between the centre $C(x - 7, 2x)$ and the point $P(-9, 11)$:
\[ \text{Distance } CP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Since $CP = \text{Radius} = 5\sqrt{2}$, we square both sides to simplify:
\[ (x_2 - x_1)^2 + (y_2 - y_1)^2 = R^2 \]

Step 3: Detailed Explanation:

• 1. Substitute the coordinates of $C(x - 7, 2x)$ and $P(-9, 11)$ into the distance equation:
\[ \left[ -9 - (x - 7) \right]^2 + (11 - 2x)^2 = (5\sqrt{2})^2 \]

• 2. Simplify the terms inside the parentheses:
- First term:
\[ -9 - x + 7 = -x - 2 = -(x + 2) \]
Squaring this yields:
\[ [-(x + 2)]^2 = (x + 2)^2 = x^2 + 4x + 4 \]
- Second term:
\[ (11 - 2x)^2 = 121 - 44x + 4x^2 \]
- Right hand side:
\[ (5\sqrt{2})^2 = 25 \times 2 = 50 \]

• 3. Combine the simplified terms into a single quadratic equation:
\[ (x^2 + 4x + 4) + (4x^2 - 44x + 121) = 50 \]
\[ 5x^2 - 40x + 125 = 50 \]

• 4. Move all terms to one side to set the quadratic equation to 0:
\[ 5x^2 - 40x + 75 = 0 \]

• 5. Divide the entire equation by 5 to simplify:
\[ x^2 - 8x + 15 = 0 \]

• 6. Solve the quadratic equation by splitting the middle term:
\[ x^2 - 5x - 3x + 15 = 0 \]
\[ x(x - 5) - 3(x - 5) = 0 \]
\[ (x - 3)(x - 5) = 0 \]

• 7. Set each factor to zero to obtain the values of $x$:
- $x - 3 = 0 \implies x = 3$
- $x - 5 = 0 \implies x = 5$


Step 4: Final Answer:
The possible values of $x$ are 3 and 5.