6.2 Q-6

Question Statement

Find the coordinates of the point of intersection of the line $x + 2y = 6$ with the circle $x^2 + y^2 - 2x - 2y - 39 = 0$.

---

Background and Explanation

To solve this problem, we need to find the points where the given line intersects the given circle. This can be done by substituting the equation of the line into the equation of the circle and solving for the values of $x$ and $y$. These values represent the coordinates of the intersection points.

Key concepts used:

• Substitution method: Solve one equation for one variable and substitute into the other equation.
• Quadratic equation: Solve for $y$ using the standard methods for solving quadratic equations (factoring in this case).
• Coordinate geometry: Find the intersection points of curves by solving systems of equations.

---

Solution

Step 1: Express $x$ in terms of $y$ from the line equation

The given line equation is:

$$x + 2y = 6 \tag{1}$$

Solving for $x$:

$$x = 6 - 2y$$

Step 2: Substitute the expression for $x$ into the circle equation

The equation of the circle is:

$$x^2 + y^2 - 2x - 2y - 39 = 0 \tag{2}$$

Substitute $x = 6 - 2y$ into equation (2):

$$(6 - 2y)^2 + y^2 - 2(6 - 2y) - 2y - 39 = 0$$

Step 3: Simplify the equation

Expand and simplify the equation step by step:

$$(6 - 2y)^2 = 36 - 24y + 4y^2$$

Substitute this into the equation:

$$36 - 24y + 4y^2 + y^2 - 12 + 4y - 2y - 39 = 0$$

Combine like terms:

$$5y^2 - 22y - 15 = 0$$

Step 4: Solve the quadratic equation

We now solve the quadratic equation:

$$5y^2 - 22y - 15 = 0$$

Factor the quadratic:

$$5y^2 - 25y - 3y - 15 = 0$$ $$5y(y - 5) + 3(y - 5) = 0$$

Factor further:

$$(5y + 3)(y - 5) = 0$$

Thus, the solutions for $y$ are:

$$y = 5 \quad \text{or} \quad y = \frac{-3}{5}$$

Step 5: Find the corresponding $x$-values

Now, substitute these $y$-values back into the equation for $x$:

• For $y = 5$:

$$x = 6 - 2(5) = 6 - 10 = -4$$

• For $y = \frac{-3}{5}$:

$$x = 6 - 2\left(\frac{-3}{5}\right) = 6 + \frac{6}{5} = \frac{30 + 6}{5} = \frac{36}{5}$$

Step 6: Final points of intersection

The points of intersection are:

$$(4, 5) \quad \text{and} \quad \left( \frac{36}{5}, \frac{-3}{5} \right)$$

---

Key Formulas or Methods Used

• Substitution method: Used to eliminate one variable by substituting the expression for one variable into the other equation.
• Quadratic equation: Solved by factoring.
• Coordinate geometry: Finding the intersection of a line and a circle.

---

Summary of Steps

1. Solve the line equation for $x$.
2. Substitute the expression for $x$ into the circle equation.
3. Simplify the resulting equation.
4. Solve the quadratic equation for $y$.
5. Substitute the values of $y$ back into the line equation to find the corresponding $x$-coordinates.
6. The points of intersection are $(4, 5)$ and $\left( \frac{36}{5}, \frac{-3}{5} \right)$.

---