Question:
If the centre of a circle \(S\) is \((2,4)\) and the length of the chord made by the secant
\[
x+y+2=0
\]
on \(S\) is \(6\) units, then the equation of the circle \(S\) is
If the centre of a circle \(S\) is \((2,4)\) and the length of the chord made by the secant
\[
x+y+2=0
\]
on \(S\) is \(6\) units, then the equation of the circle \(S\) is
Show Hint
For a chord cut by a line at distance \(d\) from the centre,
\[
\left(\frac{\text{Chord Length}}{2}\right)^2+d^2=r^2.
\]
First find the distance from the centre to the line, then compute the radius and form the circle equation.
Updated On: Jul 29, 2026
- \[ x^2+y^2-4x-8y+21=0 \]
- \[ x^2+y^2+4x+8y+21=0 \]
- \[ x^2+y^2-4x-8y-21=0 \]
- \[ x^2+y^2+4x+8y-21=0 \]
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
If a line at distance \(d\) from the centre of a circle of radius \(r\) cuts a chord of length \(L\), then
\[
L=2\sqrt{r^2-d^2}.
\]
Equivalently,
\[
r^2=d^2+\left(\frac{L}{2}\right)^2.
\]
Step 1: Find the perpendicular distance of the centre from the secant. The centre is \[ (2,4). \] The secant is \[ x+y+2=0. \] Distance from \((2,4)\) to the line is \[ d= \frac{|2+4+2|} {\sqrt{1^2+1^2}}. \] \[ = \frac{8}{\sqrt2} = 4\sqrt2. \] Hence, \[ d^2=32. \]
Step 2: Use the chord length formula. Given chord length \[ L=6. \] Therefore, \[ \frac{L}{2}=3. \] Using \[ r^2=d^2+\left(\frac{L}{2}\right)^2, \] \[ r^2=32+9. \] \[ r^2=41. \]
Step 3: Form the equation of the circle. The circle with centre \((2,4)\) and radius \(\sqrt{41}\) is \[ (x-2)^2+(y-4)^2=41. \] Expanding, \[ x^2-4x+4+y^2-8y+16=41. \] \[ x^2+y^2-4x-8y-21=0. \]
Step 4: Write the final answer. \[ \boxed{x^2+y^2-4x-8y-21=0} \]
Step 1: Find the perpendicular distance of the centre from the secant. The centre is \[ (2,4). \] The secant is \[ x+y+2=0. \] Distance from \((2,4)\) to the line is \[ d= \frac{|2+4+2|} {\sqrt{1^2+1^2}}. \] \[ = \frac{8}{\sqrt2} = 4\sqrt2. \] Hence, \[ d^2=32. \]
Step 2: Use the chord length formula. Given chord length \[ L=6. \] Therefore, \[ \frac{L}{2}=3. \] Using \[ r^2=d^2+\left(\frac{L}{2}\right)^2, \] \[ r^2=32+9. \] \[ r^2=41. \]
Step 3: Form the equation of the circle. The circle with centre \((2,4)\) and radius \(\sqrt{41}\) is \[ (x-2)^2+(y-4)^2=41. \] Expanding, \[ x^2-4x+4+y^2-8y+16=41. \] \[ x^2+y^2-4x-8y-21=0. \]
Step 4: Write the final answer. \[ \boxed{x^2+y^2-4x-8y-21=0} \]
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