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 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.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Coordinate Geometry, specifically applying the concept of distance between two points to circles.
We are given the coordinates of the center of a circle as a function of a variable $x$.
We also know a point on the circle's circumference and the radius of the circle.
Since the distance from the center to any point on the circle is equal to the radius, we can establish a distance equation.

Step 2: Key Formula or Approach:
The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by the Distance Formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
If the center is $C(x - 7, 2x)$, the point is $P(-9, 11)$, and the radius is $R = 5\sqrt{2}$, then:
\[ CP = R \implies CP^2 = R^2 \]
Squaring both sides removes the square root and simplifies the algebra.

Step 3: Detailed Explanation:

• Set up the equation using $CP^2 = R^2$:
\[ [(-9) - (x - 7)]^2 + [11 - 2x]^2 = (5\sqrt{2})^2 \]

• Simplify the expression inside the first bracket:
\[ -9 - x + 7 = -x - 2 = -(x + 2) \]
Squaring this gives:
\[ [-(x + 2)]^2 = (x + 2)^2 \]

• Now substitute this back into the equation:
\[ (x + 2)^2 + (11 - 2x)^2 = (5\sqrt{2})^2 \]

• Expand both squared terms:
\[ (x^2 + 4x + 4) + (121 - 44x + 4x^2) = 25 \times 2 \]
\[ (5x^2 - 40x + 125) = 50 \]

• Transpose 50 to the left side to form a standard quadratic equation:
\[ 5x^2 - 40x + 125 - 50 = 0 \]
\[ 5x^2 - 40x + 75 = 0 \]

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

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

• Set each factor to zero to solve for $x$:
\[ x = 3 \quad \text{or} \quad x = 5 \]


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