Question

Two circles centred at $(2,3)$ and $(4,5)$ intersect each other. If their radii are equal, then the equation of the common chord is

• A. $x + y + 1 = 0$
• B. $x + y - 1 = 0$
• C. $x + y - 7 = 0$
• D. $x + y + 7 = 0$

Hint:

When two circles have completely identical radii, their common chord is the perpendicular bisector of the line segment joining their centers! The midpoint of $C_1(2,3)$ and $C_2(4,5)$ is $\left(\frac{2+4}{2}, \frac{3+5}{2}\right) = (3,4)$. Plug $(3,4)$ into the options: $3+4-7=0$, validating option (C) immediately!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two intersecting circles with centers $C_1(2,3)$ and $C_2(4,5)$ and identical radii $r$. We need to find the Cartesian line equation representing their common chord.

Step 2: Key Formula or Approach:
The equation of the common chord between two intersecting circles $S_1 = 0$ and $S_2 = 0$ is found by subtracting one equation from the other: $$S_1 - S_2 = 0$$ Let's write down the standard geometric expanded circle configurations $(x-x_c)^2 + (y-y_c)^2 = r^2$.

Step 3: Detailed Explanation:
Let's set up the equations of the two circles with equal radius $r$: $$S_1: (x - 2)^2 + (y - 3)^2 = r^2 \implies x^2 - 4x + 4 + y^2 - 6y + 9 - r^2 = 0$$ $$S_2: (x - 4)^2 + (y - 5)^2 = r^2 \implies x^2 - 8x + 16 + y^2 - 10y + 25 - r^2 = 0$$ Subtracting equation $S_2$ from $S_1$ cancels out the quadratic $x^2, y^2$ and $r^2$ terms completely: $$S_1 - S_2 = (-4x + 8x) + (4 - 16) + (-6y + 10y) + (9 - 25) = 0$$ $$4x - 12 + 4y - 16 = 0$$ Combine the numerical constants: $$4x + 4y - 28 = 0$$ Divide the entire equation by 4 to reduce it to its final linear equation form: $$x + y - 7 = 0$$

Step 4: Final Answer:
The equation of the common chord is $x + y - 7 = 0$, which corresponds to option (C).