Precalculus Quiz: Constructing Tangents To Circles
20 questions · exam conditions
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Constructing Tangents To CirclesQuestion 1 of 20

A circle OO has center OO and radius 88. An external point PP satisfies OP=17OP=17. A tangent from PP touches the circle at TT.

For the circle described, what is the length of the tangent segment PTPT?

99
1515
2525
353\sqrt{353}
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Precalculus Quiz

Precalculus Quiz: Constructing Tangents To Circles

Practice Constructing Tangents To Circles in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Constructing Tangents To Circles, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A circle OO has center OO and radius 88. An external point PP satisfies OP=17OP=17. A tangent from PP touches the circle at TT.

For the circle described, what is the length of the tangent segment PTPT?

  1. 99
  2. 1515 (correct answer)
  3. 2525
  4. 353\sqrt{353}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 8 and the external point P is at distance d = 17 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 8² = 17², so t² + 64 = 289, thus t² = 225 and t = √225 = 15. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly reverses the Pythagorean relationship, treating the tangent length as the hypotenuse when actually the distance OP is the hypotenuse. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 2

A circle OO has center OO and radius 44. Point PP lies outside the circle with OP=5OP=5. Tangent segments from PP touch the circle at points TT and UU.

Using the given circle and point, what is the length of the tangent segment from PP to the circle (i.e., PTPT)?

  1. 33 (correct answer)
  2. 41\sqrt{41}
  3. 99
  4. 11
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 4 and the external point P is at distance d = 5 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 4² = 5², so t² + 16 = 25, thus t² = 9 and t = √9 = 3. Choice A is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 3

From point Q(0,8)Q(0, 8), tangent lines are constructed to the circle (x3)2+(y2)2=13(x-3)^2 + (y-2)^2 = 13. What is the angle between the two tangent lines?

  1. 30°30°
  2. 45°45°
  3. 60°60° (correct answer)
  4. 90°90°
Explanation: The circle has center (3,2)(3,2) and radius 13\sqrt{13}. Distance from QQ to center is 9+36=35\sqrt{9+36} = 3\sqrt{5}. In the right triangle formed by the center, external point, and tangent point, sin(α)=rd=1335=1345\sin(\alpha) = \frac{r}{d} = \frac{\sqrt{13}}{3\sqrt{5}} = \sqrt{\frac{13}{45}}. Since 1345=13450.289\frac{13}{45} = \frac{13}{45} \approx 0.289, we get sin(α)0.538\sin(\alpha) \approx 0.538, so α30°\alpha \approx 30°. The angle between the tangents is 2α=60°2\alpha = 60°.

Question 4

A circle with equation x2+y26x+4y12=0x^2 + y^2 - 6x + 4y - 12 = 0 is given. From which of the following points can exactly one tangent line be drawn to this circle?

  1. (8,2)(8, 2) only
  2. (3,2)(3, -2) only
  3. Any point at distance 55 from (3,2)(3, -2) (correct answer)
  4. Any point at distance 37\sqrt{37} from (3,2)(3, -2)
Explanation: First, rewrite the circle in standard form: (x3)2+(y+2)2=25(x-3)^2 + (y+2)^2 = 25, so center is (3,2)(3,-2) and radius is 55. Exactly one tangent can be drawn from a point if and only if the point lies on the circle itself. Points on the circle are at distance 55 from the center (3,2)(3,-2). Choice A is a specific external point, choice B is the center, and choice D represents points outside the circle.

Question 5

A circle OO has center OO and radius 55. An external point PP is such that OP=13OP=13. From PP, a tangent segment PTPT is drawn to the circle, touching the circle at TT.

For the circle described, what is the length of the tangent segment PTPT?

  1. 88
  2. 1212 (correct answer)
  3. 1818
  4. 194\sqrt{194}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 5 and the external point P is at distance d = 13 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 5² = 13², so t² + 25 = 169, thus t² = 144 and t = √144 = 12. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice A incorrectly reverses the Pythagorean relationship, treating the tangent length as the hypotenuse when actually the distance OP is the hypotenuse. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 6

Circle OO has center OO and radius 88. Point PP is outside the circle with OP=17OP=17. Tangent segments PTPT and PUPU are drawn from PP to the circle, touching at TT and UU.

Which of the following is true about the two tangent segments from PP?

  1. PT<PUPT < PU
  2. PT=PUPT = PU (correct answer)
  3. PT>PUPT > PU
  4. One of PTPT and PUPU must be a diameter of the circle
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d, same radius r), both have length t = √(d² - r²), making them equal; additionally, the configuration is symmetric about line OP. Choice B is correct because it correctly identifies equal lengths. Choice A incorrectly claims the two tangent segments have different lengths, but by symmetry and the Pythagorean theorem with the same parameters, they must be equal. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 7

A circle OO has center OO and radius 66. An external point PP satisfies OP=10OP=10. A tangent from PP touches the circle at TT.

For the circle described, what is the length of the tangent segment PTPT?

  1. 1616
  2. 88 (correct answer)
  3. 44
  4. 136\sqrt{136}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (tt), radius (rr), and distance from center to external point (dd) forms a right triangle where t2+r2=d2t^2 + r^2 = d^2 by the Pythagorean theorem. Given that the circle has radius r=6r = 6 and the external point PP is at distance d=10d = 10 from center OO, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OPOP as the hypotenuse, giving us t2+62=102t^2 + 6^2 = 10^2, so t2+36=100t^2 + 36 = 100, thus t2=64t^2 = 64 and t=64=8t = \sqrt{64} = 8. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice A incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: dr+td \neq r + t, rather d2=r2+t2d^2 = r^2 + t^2. Key to tangent problems: remember that tangent \perp radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius rr and distance dd to external point, then use t=d2r2t = \sqrt{d^2 - r^2}; if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 8

Circle OO has radius 33, and an external point PP satisfies OP=5OP=5. Tangent segments from PP touch the circle at points of tangency. Using the given circle and point, how many tangent lines can be drawn from point PP to circle OO?​​​

  1. 00
  2. 11
  3. 22 (correct answer)
  4. Infinitely many
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. From external point P, we can construct exactly two tangent lines because as we rotate a line through P, there are exactly two positions where the line is perpendicular to a radius, and those positions define the two tangent lines. Choice C is correct because it correctly counts tangents. Choice B incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 9

Circle OO has center OO and radius 44. Point PP is outside the circle, and a tangent segment PTPT touches the circle at TT. To construct a tangent from PP to circle OO, what must be true at the point of tangency TT?​​

  1. The segment OTOT is perpendicular to the tangent line PTPT. (correct answer)
  2. The segment OTOT is parallel to the tangent line PTPT.
  3. The point TT must lie on line OPOP.
  4. The segment PTPT must pass through the center OO.
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. At the point of tangency, the radius and tangent line meet at a right angle (90°), which is the defining property of a tangent line. The tangent line must be perpendicular to the radius OT at point T because this is the definition of tangency—any line through T that is not perpendicular to OT would either miss the circle entirely or intersect it at a second point. Choice A is correct because it correctly states the perpendicularity property that defines tangency. Choice B incorrectly claims the radius and tangent are parallel, but the defining property of tangency is perpendicularity. When constructing tangents, the perpendicularity condition is essential: the tangent line must be perpendicular to the radius at the point where it touches the circle.

Question 10

A circle is tangent to both coordinate axes in the first quadrant. If a tangent line from point (8,0)(8, 0) to this circle has slope m=34m = -\frac{3}{4}, what is the radius of the circle?

  1. 22
  2. 33 (correct answer)
  3. 44
  4. 55
Explanation: Since the circle is tangent to both axes in the first quadrant, its center is at (r,r)(r,r) where rr is the radius. The tangent line from (8,0)(8,0) with slope 34-\frac{3}{4} has equation y=34(x8)=34x+6y = -\frac{3}{4}(x-8) = -\frac{3}{4}x + 6. The distance from center (r,r)(r,r) to this line must equal the radius: 34r+r6916+1=r\frac{|\frac{3}{4}r + r - 6|}{\sqrt{\frac{9}{16} + 1}} = r. This gives 74r654=r\frac{|\frac{7}{4}r - 6|}{\frac{5}{4}} = r, so 74r6=54r|\frac{7}{4}r - 6| = \frac{5}{4}r. Since r<6r < 6, we have 674r=54r6 - \frac{7}{4}r = \frac{5}{4}r, giving 6=3r6 = 3r, so r=3r = 3.

Question 11

Circle OO has center OO and radius 44. Point PP is an external point with OP=5OP=5. Two tangent segments from PP touch the circle at T1T_1 and T2T_2. For the circle described, which of the following is true about the two tangent segments PT1PT_1 and PT2PT_2?

  1. PT1=PT2PT_1 = PT_2 (correct answer)
  2. PT1>PT2PT_1 > PT_2
  3. PT1<PT2PT_1 < PT_2
  4. There is only one tangent segment from PP
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d = 5, same radius r = 4), both have length t = √(d² - r²) = √(25 - 16) = √9 = 3, making them equal; additionally, the configuration is symmetric about line OP. Choice A is correct because it correctly identifies equal lengths. Choice D incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration.

Question 12

A circle OO has center O(0,0)O(0,0) and radius 55. Point PP is located at (13,0)(13,0), and a tangent segment PTPT is drawn from PP to the circle, touching the circle at point TT. For the circle described, what is the length of the tangent segment PTPT?

  1. 1818
  2. 1212 (correct answer)
  3. 88
  4. ext13 ext{13}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. A tangent line to a circle is a line that touches the circle at exactly one point, called the point of tangency, and is perpendicular to the radius at that point. Given that the circle has radius r = 5 and the external point P is at distance d = 13 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 5² = 13², so t² + 25 = 169, thus t² = 144 and t = √144 = 12. Choice B is correct because it applies the Pythagorean theorem correctly with the given values. Choice D incorrectly uses 13 as the tangent length, but 13 is actually the distance from O to P, not the tangent length. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

Question 13

Circle OO has center OO and radius 44. An external point PP satisfies OP=10OP=10, and tangents from PP touch the circle at TT and UU. For the circle described, which of the following is true about the two tangent segments PTPT and PUPU?​​​

  1. PT>PUPT > PU
  2. PT<PUPT < PU
  3. PT=PUPT = PU (correct answer)
  4. Their lengths cannot be compared without coordinates.
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. Because both tangent segments from P satisfy the same right triangle relationship (same distance d, same radius r), both have length t = √(d² - r²), making them equal; additionally, the configuration is symmetric about line OP. Choice C is correct because it correctly identifies equal lengths. Choice D incorrectly claims their lengths cannot be compared without coordinates, but by symmetry and the Pythagorean theorem with the same parameters, they must be equal. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

Question 14

Circle OO has center OO and radius 66. Point PP is an external point with OP=10OP=10. Two tangent lines can be drawn from PP to the circle.

For the circle described, how many tangent lines can be drawn from point PP to circle OO?

  1. 00
  2. 11
  3. 22 (correct answer)
  4. Infinitely many
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. From a point outside a circle, exactly two tangent lines can be drawn to the circle, and the two tangent segments (from the external point to the points of tangency) have equal length. From external point P, we can construct exactly two tangent lines because as we rotate a line through P, there are exactly two positions where the line is perpendicular to a radius, and those positions define the two tangent lines. Choice C is correct because it correctly counts tangents. Choice B incorrectly states there is only one tangent from an external point, but there are always exactly two tangent lines from any external point. From any external point, exactly two tangent lines can be drawn to a circle, and the two tangent segments are always equal in length due to the symmetry of the configuration. Remember: external point → 2 tangents; point on circle → 1 tangent; interior point → 0 tangents (no real tangent lines can be drawn).

Question 15

A circle OO has center OO and radius 77. An external point PP satisfies OP=25OP=25. A tangent segment PTPT touches the circle at TT.

For the circle described, what is the length of the tangent segment PTPT?

  1. 2424 (correct answer)
  2. 1818
  3. 3232
  4. 674\sqrt{674}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. Choice A is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and tangent length instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 16

Circle OO has center OO and radius 77. Point PP is an external point with OP=25OP=25, and a tangent segment PTPT touches the circle at TT. For the circle described, what is the length of the tangent segment PTPT?

  1. 1818
  2. 2424 (correct answer)
  3. 2626
  4. 3232
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly uses 26, which would be the hypotenuse if the tangent were 24 and radius were 10, mixing up the problem parameters. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 17

Circle OO has radius 44, and an external point PP satisfies OP=5OP=5. Tangent segments from PP touch the circle at TT and TT'. Using the given circle and point, what is the length of the tangent segment PTPT?

  1. 11
  2. 33 (correct answer)
  3. 99
  4. 41\sqrt{41}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 4 and the external point P is at distance d = 5 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 4² = 5², so t² + 16 = 25, thus t² = 9 and t = √9 = 3. This is a 3-4-5 Pythagorean triple, so we immediately recognize the tangent length is 3. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and distance instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 18

Circle OO has center OO and radius 77. Point PP is an external point with OP=25OP=25. Tangent segment PTPT touches the circle at TT. For the circle described, what is the length of the tangent segment PTPT?

  1. 1818
  2. 2424 (correct answer)
  3. 2626
  4. 674\sqrt{674}
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. The relationship between the tangent length (t), radius (r), and distance from center to external point (d) forms a right triangle where t² + r² = d² by the Pythagorean theorem. Given that the circle has radius r = 7 and the external point P is at distance d = 25 from center O, we form a right triangle with the radius as one leg, the tangent segment as the other leg, and OP as the hypotenuse, giving us t² + 7² = 25², so t² + 49 = 625, thus t² = 576 and t = √576 = 24. This is a 7-24-25 Pythagorean triple, so we immediately recognize the tangent length is 24. Choice B is correct because it applies the Pythagorean theorem correctly with specific values. Choice C incorrectly adds the radius and distance instead of using the Pythagorean theorem: d ≠ r + t, rather d² = r² + t². Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length. To find tangent length from external point, identify radius r and distance d to external point, then use t = √(d² - r²); if you recognize a Pythagorean triple, you can determine the answer immediately.

Question 19

Circle OO has center OO and radius 55. Point PP is an external point, and a tangent from PP touches the circle at TT.

To construct a tangent from PP to circle OO, what must be true at the point of tangency TT?

  1. The segment OTOT is perpendicular to the line PTPT (correct answer)
  2. The segment OTOT is parallel to the line PTPT
  3. The segment PTPT must pass through the center OO
  4. The point TT must be the midpoint of segment OPOP
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. A tangent line to a circle is a line that touches the circle at exactly one point, called the point of tangency, and is perpendicular to the radius at that point. The tangent line must be perpendicular to the radius OT at point T because this is the definition of tangency—any line through T that is not perpendicular to OT would either miss the circle entirely or intersect it at a second point. Choice A is correct because it correctly states the perpendicularity property. Choice B claims the tangent and radius are parallel or at an angle other than 90°, but the defining property of tangency is perpendicularity. When constructing tangents, the perpendicularity condition is essential: the tangent line must be perpendicular to the radius at the point where it touches the circle. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.

Question 20

Circle OO has center OO and radius 1010. Point PP is outside the circle, and a tangent from PP touches the circle at TT.

For the circle described, what is the angle between the radius OTOT and the tangent line PTPT at the point of tangency TT?

  1. 6060^\circ
  2. 9090^\circ (correct answer)
  3. 120120^\circ
  4. 180180^\circ
Explanation: This question tests understanding of tangent lines from an external point to a circle and their geometric properties. At the point of tangency, the radius and tangent line meet at a right angle (90°), which is the defining property of a tangent line. The tangent line must be perpendicular to the radius OT at point T because this is the definition of tangency—any line through T that is not perpendicular to OT would either miss the circle entirely or intersect it at a second point. Choice B is correct because it correctly states the perpendicularity property. Choice A claims the tangent and radius are parallel or at an angle other than 90°, but the defining property of tangency is perpendicularity. When constructing tangents, the perpendicularity condition is essential: the tangent line must be perpendicular to the radius at the point where it touches the circle. Key to tangent problems: remember that tangent ⊥ radius at point of tangency creates a right triangle with the segment from center to external point as hypotenuse, allowing use of the Pythagorean theorem to find tangent length.