Question

The lines \(2x-3y=5\) and \(3x-4y=7\) are the diameters of a circle of area \(154\) sq.unit. The equation of this circle is

• A. \(x^2+y^2+2x-2y=62\)
• B. \(x^2+y^2+2x-2y=47\)
• C. \(x^2+y^2-2x+2y=47\)
• D. \(x^2+y^2-2x+2y=62\)

Hint:

If two lines are diameters of a circle, their point of intersection gives the center of the circle.

Answer 1

The Correct Option is B

Concept:
A diameter of a circle always passes through the center of the circle. If two lines are diameters of the same circle, then their intersection point is the center of the circle. The standard equation of a circle with center \((h,k)\) and radius \(r\) is: \[ (x-h)^2+(y-k)^2=r^2 \]

Step 1: Find the center of the circle.
The two diameters are: \[ 2x-3y=5 \] and \[ 3x-4y=7 \] The center is the intersection of these two lines. Solve: \[ 2x-3y=5 \] \[ 3x-4y=7 \] Multiply first equation by \(3\): \[ 6x-9y=15 \] Multiply second equation by \(2\): \[ 6x-8y=14 \] Subtract: \[ (6x-9y)-(6x-8y)=15-14 \] \[ -y=1 \] \[ y=-1 \] Substitute in: \[ 2x-3y=5 \] \[ 2x-3(-1)=5 \] \[ 2x+3=5 \] \[ 2x=2 \] \[ x=1 \] So the center is: \[ (1,-1) \]

Step 2: Find the radius from area.
Given area of circle: \[ 154 \] Area formula: \[ \pi r^2=154 \] Using: \[ \pi=\frac{22}{7} \] \[ \frac{22}{7}r^2=154 \] \[ r^2=\frac{154\times7}{22} \] \[ r^2=49 \]

Step 3: Write the circle equation.
Center: \[ (h,k)=(1,-1) \] So: \[ (x-1)^2+(y+1)^2=49 \]

Step 4: Expand the equation.
\[ (x-1)^2=x^2-2x+1 \] \[ (y+1)^2=y^2+2y+1 \] So: \[ x^2-2x+1+y^2+2y+1=49 \] \[ x^2+y^2-2x+2y+2=49 \] \[ x^2+y^2-2x+2y=47 \]

Step 5: Check with options.
The obtained equation is: \[ x^2+y^2-2x+2y=47 \] This corresponds to option (C), not option (B), if the center is \((1,-1)\). Hence, mathematically the correct answer is: \[ \boxed{(C)\ x^2+y^2-2x+2y=47} \]