Question:medium
The area of the circle \( x^2 - 2x + y^2 - 10y + k = 0 \) is \( 25\pi \). The value of \( k \) is equal to:
The area of the circle \( x^2 - 2x + y^2 - 10y + k = 0 \) is \( 25\pi \). The value of \( k \) is equal to:
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Completing the square is an alternative: \( (x^2 - 2x + 1) + (y^2 - 10y + 25) = 1 + 25 - k \). This gives \( (x-1)^2 + (y-5)^2 = 26 - k \). Since \( r^2 = 25 \), we have \( 26 - k = 25 \), so \( k = 1 \).
Updated On: May 1, 2026
- \( -1 \)
- \( 1 \)
- \( 0 \)
- \( 2 \)
- \( 3 \)
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The Correct Option is B
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