Geometry and Trigonometry practice question

Answer: A. 2

Completing the square: x² + 10x + y² − 6y = 2 → (x + 5)² + (y − 3)² = 2 + 25 + 9 = 36, so the center is C(−5, 3) and the radius is 6. A tangent line is perpendicular to the radius at the point of tangency, so triangle PTC is a right triangle with legs PT and TC = 6 and hypotenuse PC. Here PC² = (1 − (−5))² + (1 − 3)² = 36 + 4 = 40, so PT = √(40 − 36) = √4 = 2. (B) comes from a sign error when completing the square — moving the −2 to the right side as −2 instead of +2, giving r² = 32 and √(40 − 32) = 2√2. (C) comes from subtracting the radius from the distance to the center (2√10 − 6) rather than using the Pythagorean relationship; a tangent length is a leg, not a difference of lengths. (D) is the distance PC = √40 = 2√10 itself, i.e., forgetting to subtract r² and treating the distance to the center as the tangent length.

Why the other answers are wrong

B. 2√2

When you moved the −2 across the equal sign you kept it negative, getting r² = 25 + 9 − 2 = 32 instead of 36. The constant becomes +2, so r² = 36 and PT = √(40 − 36) = 2.

C. 2√10 − 6

You subtracted the radius from the distance PC (2√10 − 6), but the tangent length is a leg of a right triangle, not a difference of lengths — use PT = √(PC² − r²) = √(40 − 36) = 2.

D. 2√10

√40 = 2√10 is the distance PC from P to the center, not the tangent length. Subtract r² first: PT = √(40 − 36) = 2.

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