Question:medium

Let \( p(x, y) \) be a variable point on the circle \( x^2 + y^2 - 6x - 8y + 21 = 0 \). Then the maximum possible distance from the vertex of \( y^2 + 6y + x + 13 = 0 \) is:

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To find the vertex of a parabola quickly, differentiate the squared variable part. Here, \( \frac{d}{dy}(y^2 + 6y) = 2y + 6 = 0 \implies y = -3 \). Then substitute back to find \( x \).

Updated On: Apr 6, 2026

\( 7 + 2\sqrt{2} \)
\( 2 + 7\sqrt{2} \)
\( 4 + 7\sqrt{2} \)
\( 3 + 2\sqrt{2} \)

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The Correct Option is B

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Let \( p(x, y) \) be a variable point on the circle \( x^2 + y^2 - 6x - 8y + 21 = 0 \). Then the maximum possible distance from the vertex of \( y^2 + 6y + x + 13 = 0 \) is:

Show Hint

The Correct Option is B

Solution and Explanation

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