Question:
If the points
\[
(2,0),\ (0,1),\ (4,0),\ (0,k)
\]
are concyclic, then \(k=\)
If the points
\[
(2,0),\ (0,1),\ (4,0),\ (0,k)
\]
are concyclic, then \(k=\)
Show Hint
When four points are concyclic, first determine the circle using three points. Then substitute the fourth point into the circle equation to obtain the required parameter.
Updated On: Jul 29, 2026
- \(4\)
- \(6\)
- \(8\)
- \(2\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
The general equation of a circle is
\[
x^2+y^2+2gx+2fy+c=0.
\]
Since all points on the circle satisfy this equation, substitute the given points to determine the circle and then find \(k\).
Step 1: Substitute the point \((2,0)\). \[ 2^2+0^2+4g+c=0. \] \[ 4+4g+c=0. \] \[ 4g+c=-4. \] \[ \cdots (1) \]
Step 2: Substitute the point \((4,0)\). \[ 4^2+0^2+8g+c=0. \] \[ 16+8g+c=0. \] \[ 8g+c=-16. \] \[ \cdots (2) \] Subtracting (1) from (2), \[ 4g=-12. \] \[ g=-3. \] Substituting in (1), \[ c=8. \]
Step 3: Substitute the point \((0,1)\). \[ 0^2+1^2+2f+8=0. \] \[ 1+2f+8=0. \] \[ 2f=-9. \] \[ f=-\frac92. \] Hence the circle is \[ x^2+y^2-6x-9y+8=0. \]
Step 4: Substitute the point \((0,k)\). Since \((0,k)\) lies on the circle, \[ k^2-9k+8=0. \] Factorizing, \[ (k-1)(k-8)=0. \] \[ k=1 \quad \text{or} \quad k=8. \]
Step 5: Choose the valid value. The point \((0,1)\) is already one of the given points. For four distinct concyclic points, \[ k\neq 1. \] Therefore, \[ k=8. \]
Step 6: Write the final answer. \[ \boxed{8} \]
Step 1: Substitute the point \((2,0)\). \[ 2^2+0^2+4g+c=0. \] \[ 4+4g+c=0. \] \[ 4g+c=-4. \] \[ \cdots (1) \]
Step 2: Substitute the point \((4,0)\). \[ 4^2+0^2+8g+c=0. \] \[ 16+8g+c=0. \] \[ 8g+c=-16. \] \[ \cdots (2) \] Subtracting (1) from (2), \[ 4g=-12. \] \[ g=-3. \] Substituting in (1), \[ c=8. \]
Step 3: Substitute the point \((0,1)\). \[ 0^2+1^2+2f+8=0. \] \[ 1+2f+8=0. \] \[ 2f=-9. \] \[ f=-\frac92. \] Hence the circle is \[ x^2+y^2-6x-9y+8=0. \]
Step 4: Substitute the point \((0,k)\). Since \((0,k)\) lies on the circle, \[ k^2-9k+8=0. \] Factorizing, \[ (k-1)(k-8)=0. \] \[ k=1 \quad \text{or} \quad k=8. \]
Step 5: Choose the valid value. The point \((0,1)\) is already one of the given points. For four distinct concyclic points, \[ k\neq 1. \] Therefore, \[ k=8. \]
Step 6: Write the final answer. \[ \boxed{8} \]
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