Question:
If the segments of the straight lines
\[
x+y=6
\]
and
\[
x+2y=4
\]
are two diameters of a circle passing through \((6,2)\), then the equation of that circle is:
If the segments of the straight lines
\[
x+y=6
\]
and
\[
x+2y=4
\]
are two diameters of a circle passing through \((6,2)\), then the equation of that circle is:
Show Hint
If two lines are diameters of a circle, their point of intersection is the centre of the circle.
Updated On: Jun 18, 2026
- \(x^2+y^2-2x-4y-20=0\)
- \(x^2+y^2+6x-4y-68=0\)
- \(x^2+y^2-16x+4y+48=0\)
- \(x^2+y^2+2x-10y-32=0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Find the centre of the circle.
Since the two given lines are diameters of the circle, their point of intersection is the centre of the circle.
The equations are \[ x+y=6 \] and \[ x+2y=4. \] Subtracting the first equation from the second, \[ y=-2. \] Substituting in \[ x+y=6, \] we get \[ x-2=6. \] \[ x=8. \] So, the centre is \[ (8,-2). \]
Step 2: Find the radius.
The circle passes through \[ (6,2). \] Therefore, the radius is the distance between \((8,-2)\) and \((6,2)\). \[ r^2=(6-8)^2+(2+2)^2. \] \[ r^2=(-2)^2+4^2. \] \[ r^2=4+16=20. \]
Step 3: Write the equation of the circle.
The equation of a circle with centre \((8,-2)\) and radius squared \(20\) is \[ (x-8)^2+(y+2)^2=20. \] Expanding, \[ x^2-16x+64+y^2+4y+4=20. \] \[ x^2+y^2-16x+4y+68-20=0. \] \[ x^2+y^2-16x+4y+48=0. \]
Step 4: Final conclusion.
Therefore, the required equation is \[ \boxed{x^2+y^2-16x+4y+48=0} \]
Since the two given lines are diameters of the circle, their point of intersection is the centre of the circle.
The equations are \[ x+y=6 \] and \[ x+2y=4. \] Subtracting the first equation from the second, \[ y=-2. \] Substituting in \[ x+y=6, \] we get \[ x-2=6. \] \[ x=8. \] So, the centre is \[ (8,-2). \]
Step 2: Find the radius.
The circle passes through \[ (6,2). \] Therefore, the radius is the distance between \((8,-2)\) and \((6,2)\). \[ r^2=(6-8)^2+(2+2)^2. \] \[ r^2=(-2)^2+4^2. \] \[ r^2=4+16=20. \]
Step 3: Write the equation of the circle.
The equation of a circle with centre \((8,-2)\) and radius squared \(20\) is \[ (x-8)^2+(y+2)^2=20. \] Expanding, \[ x^2-16x+64+y^2+4y+4=20. \] \[ x^2+y^2-16x+4y+68-20=0. \] \[ x^2+y^2-16x+4y+48=0. \]
Step 4: Final conclusion.
Therefore, the required equation is \[ \boxed{x^2+y^2-16x+4y+48=0} \]
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