Question:
The centre and radius of the circle $x^2 + y^2 - 2x + 4y = 8$ respectively are:
The centre and radius of the circle $x^2 + y^2 - 2x + 4y = 8$ respectively are:
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Always complete squares carefully to avoid sign errors.
Updated On: Apr 30, 2026
- \((1,2), \sqrt{13} \)
- \((-1,2), \sqrt{13} \)
- \((-1,-1), \sqrt{13} \)
- \((1,-2), \sqrt{13} \)
- \((2,1), \sqrt{13} \)
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The Correct Option is D
Solution and Explanation
Concept:
Standard form:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Step 1: Complete squares.
\[ (x^2 - 2x) + (y^2 + 4y) = 8 \] \[ (x-1)^2 -1 + (y+2)^2 -4 = 8 \] \[ (x-1)^2 + (y+2)^2 = 13 \]
Step 2: Identify parameters.
Centre: \[ (1,-2) \] Radius: \[ \sqrt{13} \]
Step 1: Complete squares.
\[ (x^2 - 2x) + (y^2 + 4y) = 8 \] \[ (x-1)^2 -1 + (y+2)^2 -4 = 8 \] \[ (x-1)^2 + (y+2)^2 = 13 \]
Step 2: Identify parameters.
Centre: \[ (1,-2) \] Radius: \[ \sqrt{13} \]
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