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:

Squaring both sides of the distance formula immediately helps avoid working with messy square root signs.

Answer 1

Given:
Centre of the circle = \((x - 7,\ 2x)\)
Circle passes through point \((-9,\ 11)\)
Radius = \(5\sqrt{2}\)

Step 1: Use distance formula
Distance between centre \((x - 7,\ 2x)\) and point \((-9, 11)\) must equal radius:
\[ \sqrt{(x - 7 + 9)^2 + (2x - 11)^2} = 5\sqrt{2} \]
Simplify inside:
\[ (x + 2)^2 + (2x - 11)^2 = (5\sqrt{2})^2 \] \[ (x + 2)^2 + (2x - 11)^2 = 50 \]

Step 2: Expand the squares
\[ (x + 2)^2 = x^2 + 4x + 4 \] \[ (2x - 11)^2 = 4x^2 - 44x + 121 \]
Add them: \[ x^2 + 4x + 4 + 4x^2 - 44x + 121 = 50 \] \[ 5x^2 - 40x + 125 = 50 \]

Step 3: Solve the quadratic
\[ 5x^2 - 40x + 75 = 0 \] Divide entire equation by 5: \[ x^2 - 8x + 15 = 0 \] Factorise: \[ (x - 3)(x - 5) = 0 \]
So, \[ x = 3 \quad \text{or} \quad x = 5 \]

Final Answer:
The possible values of \(x\) are:
\[ \boxed{3 \text{ and } 5} \]