Question

If two diameters of a circle are along the lines $2x-3y=5$ and $3x-4y=7$, then the centre is at

• A. (1, 1)
• B. (-1, 1)
• C. (-1, -1)
• D. (1, -1)
• E. (1, -2)

Hint:

Intersection of any two diameters always gives the coordinates of the center.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
All diameters of a circle must pass through its center. Therefore, the center of the circle is the point of intersection of any two of its diameters.
Step 2: Key Formula or Approach:
To find the center, we need to solve the system of two linear equations representing the two given diameters.
Equation 1: \( 2x - 3y = 5 \)
Equation 2: \( 3x - 4y = 7 \)
Step 3: Detailed Explanation:
We can solve this system using the elimination method. Let's eliminate the variable x.
Multiply Equation 1 by 3:
\[ 3(2x - 3y) = 3(5) \implies 6x - 9y = 15 \] Multiply Equation 2 by 2:
\[ 2(3x - 4y) = 2(7) \implies 6x - 8y = 14 \] Now, subtract the new second equation from the new first equation:
\[ (6x - 9y) - (6x - 8y) = 15 - 14 \] \[ 6x - 9y - 6x + 8y = 1 \] \[ -y = 1 \implies y = -1 \] Substitute the value of \( y = -1 \) back into one of the original equations to find x. Let's use Equation 1:
\[ 2x - 3(-1) = 5 \] \[ 2x + 3 = 5 \] \[ 2x = 5 - 3 \] \[ 2x = 2 \implies x = 1 \] The point of intersection is (1, -1).
Step 4: Final Answer:
The centre of the circle is at (1, -1).