Question:
Segments of lines
\[
2x+3y=1
\]
and
\[
4x-3y=11
\]
are diameters of a circle of area \(153.94\) square units. Then the equation of circle with integer radius is:
Segments of lines
\[
2x+3y=1
\]
and
\[
4x-3y=11
\]
are diameters of a circle of area \(153.94\) square units. Then the equation of circle with integer radius is:
Show Hint
If diameters are given as lines, their point of intersection is the centre of the circle.
Updated On: Jun 11, 2026
- \(x^2+y^2+4x-2y-44=0\)
- \(x^2+y^2-4x+2y+44=0\)
- \(x^2+y^2-4x+2y-44=0\)
- \(x^2+y^2+4x-2y+44=0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Find the centre of the circle.
The diameters intersect at the centre. Solving \[ 2x+3y=1 \] and \[ 4x-3y=11, \] adding, \[ 6x=12 \] \[ x=2. \] Substituting, \[ 4+3y=1 \] \[ y=-1. \] Thus centre is \[ (2,-1). \]
Step 2: Find the radius.
Area \[ =\pi r^2 = 153.94. \] Using \[ \pi=\frac{22}{7}, \] \[ r^2 = 153.94\times\frac7{22} = 49. \] Hence \[ r=7. \]
Step 3: Form the equation.
\[ (x-2)^2+(y+1)^2=49. \] Expanding, \[ x^2+y^2-4x+2y+5=49. \] \[ x^2+y^2-4x+2y-44=0. \] \[ \boxed{x^2+y^2-4x+2y-44=0} \]
The diameters intersect at the centre. Solving \[ 2x+3y=1 \] and \[ 4x-3y=11, \] adding, \[ 6x=12 \] \[ x=2. \] Substituting, \[ 4+3y=1 \] \[ y=-1. \] Thus centre is \[ (2,-1). \]
Step 2: Find the radius.
Area \[ =\pi r^2 = 153.94. \] Using \[ \pi=\frac{22}{7}, \] \[ r^2 = 153.94\times\frac7{22} = 49. \] Hence \[ r=7. \]
Step 3: Form the equation.
\[ (x-2)^2+(y+1)^2=49. \] Expanding, \[ x^2+y^2-4x+2y+5=49. \] \[ x^2+y^2-4x+2y-44=0. \] \[ \boxed{x^2+y^2-4x+2y-44=0} \]
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