Question:
Suppose a circle passes through \((0,a)\) and \((b,h)\) having its centre at \((c,0)\). Then the value of \(c\) is
Suppose a circle passes through \((0,a)\) and \((b,h)\) having its centre at \((c,0)\). Then the value of \(c\) is
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For any circle, all points on the circle are equidistant from the centre. Equate the distance formulas from the centre to different points on the circle to find unknown parameters.
Updated On: Jun 24, 2026
- \(\dfrac{b^2-a^2+h^2}{2b}\)
- \(\dfrac{b^2+a^2-h^2}{2b}\)
- \(\dfrac{b^2-a^2+h^2}{2a}\)
- \(\dfrac{b^2+a^2-h^2}{2a}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the property of equal radii.
Since the circle passes through \((0,a)\) and \((b,h)\) and has centre \((c,0)\), the distances from the centre to these points are equal.
Thus, \[ \sqrt{(0-c)^2+(a-0)^2} = \sqrt{(b-c)^2+h^2} \]
Step 2: Square both sides.
\[ c^2+a^2 = (b-c)^2+h^2 \] Expanding, \[ c^2+a^2 = b^2-2bc+c^2+h^2 \] Cancelling \(c^2\), \[ a^2=b^2-2bc+h^2 \]
Step 3: Solve for \(c\).
\[ 2bc=b^2-a^2+h^2 \] Therefore, \[ c=\frac{b^2-a^2+h^2}{2b} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{b^2-a^2+h^2}{2b}} \]
Since the circle passes through \((0,a)\) and \((b,h)\) and has centre \((c,0)\), the distances from the centre to these points are equal.
Thus, \[ \sqrt{(0-c)^2+(a-0)^2} = \sqrt{(b-c)^2+h^2} \]
Step 2: Square both sides.
\[ c^2+a^2 = (b-c)^2+h^2 \] Expanding, \[ c^2+a^2 = b^2-2bc+c^2+h^2 \] Cancelling \(c^2\), \[ a^2=b^2-2bc+h^2 \]
Step 3: Solve for \(c\).
\[ 2bc=b^2-a^2+h^2 \] Therefore, \[ c=\frac{b^2-a^2+h^2}{2b} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{b^2-a^2+h^2}{2b}} \]
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