Question:
The area of the circle passing through the points \((5,\pm 2)\), \((1,2)\) is
The area of the circle passing through the points \((5,\pm 2)\), \((1,2)\) is
Show Hint
The center of a circle can be found by intersecting the perpendicular bisectors of any two chords of the circle.
Updated On: Jun 22, 2026
- \(8\pi\)
- \(4\pi\)
- \(2\pi\)
- \(16\pi\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the given points.
The given points are
\[ (5,2),\quad (5,-2),\quad (1,2) \]
Step 2: Find the perpendicular bisector of the vertical chord.
The points \((5,2)\) and \((5,-2)\) form a vertical chord.
Its midpoint is
\[ \left(5,\frac{2+(-2)}{2}\right)=(5,0) \] Therefore, the perpendicular bisector is the horizontal line
\[ y=0 \]
Step 3: Find the perpendicular bisector of another chord.
Consider the points \((5,2)\) and \((1,2)\).
This is a horizontal chord.
Its midpoint is
\[ \left(\frac{5+1}{2},\frac{2+2}{2}\right)=(3,2) \] Therefore, its perpendicular bisector is the vertical line
\[ x=3 \]
Step 4: Find the center of the circle.
The center is the intersection of the two perpendicular bisectors:
\[ x=3,\quad y=0 \] Hence, the center is
\[ (3,0) \]
Step 5: Find the radius of the circle.
Using the distance formula between the center \((3,0)\) and the point \((5,2)\),
\[ r=\sqrt{(5-3)^2+(2-0)^2} \] \[ =\sqrt{2^2+2^2} \] \[ =\sqrt{4+4} \] \[ =\sqrt{8} \] Thus,
\[ r^2=8 \]
Step 6: Calculate the area of the circle.
Area of a circle is
\[ \pi r^2 \] Therefore,
\[ \text{Area}=\pi(8) \] \[ =8\pi \]
Step 7: Final conclusion.
Hence, the area of the circle is
\[ \boxed{8\pi} \]
The given points are
\[ (5,2),\quad (5,-2),\quad (1,2) \]
Step 2: Find the perpendicular bisector of the vertical chord.
The points \((5,2)\) and \((5,-2)\) form a vertical chord.
Its midpoint is
\[ \left(5,\frac{2+(-2)}{2}\right)=(5,0) \] Therefore, the perpendicular bisector is the horizontal line
\[ y=0 \]
Step 3: Find the perpendicular bisector of another chord.
Consider the points \((5,2)\) and \((1,2)\).
This is a horizontal chord.
Its midpoint is
\[ \left(\frac{5+1}{2},\frac{2+2}{2}\right)=(3,2) \] Therefore, its perpendicular bisector is the vertical line
\[ x=3 \]
Step 4: Find the center of the circle.
The center is the intersection of the two perpendicular bisectors:
\[ x=3,\quad y=0 \] Hence, the center is
\[ (3,0) \]
Step 5: Find the radius of the circle.
Using the distance formula between the center \((3,0)\) and the point \((5,2)\),
\[ r=\sqrt{(5-3)^2+(2-0)^2} \] \[ =\sqrt{2^2+2^2} \] \[ =\sqrt{4+4} \] \[ =\sqrt{8} \] Thus,
\[ r^2=8 \]
Step 6: Calculate the area of the circle.
Area of a circle is
\[ \pi r^2 \] Therefore,
\[ \text{Area}=\pi(8) \] \[ =8\pi \]
Step 7: Final conclusion.
Hence, the area of the circle is
\[ \boxed{8\pi} \]
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