math-2017:G.C.4
math-2017:G.C.4 Construct a tangent line from a point outside a given circle to the circle.
Learning goal: Construct, with compass and straightedge, a tangent line from a point outside a given circle to the circle, and prove that the constructed line is tangent using the inscribed angle on a diameter and the perpendicularity of a tangent and a radius.
Scope note. G.C.4 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full as part of the Geometry course, as extension content for students on that path.
The problem. A circle with center O is given, and a point P outside it. We want a line through P that touches the circle at exactly one point. Guessing with a ruler is not a construction: the line that "looks tangent" is almost always a secant that cuts the circle in two very close points or misses it. A construction uses only a compass and a straightedge, and comes with a proof that the result is exactly what was wanted.
What we know about tangents. From the previous lesson: a line through a point T of the circle is tangent at T if and only if it is perpendicular to the radius OT. So the whole problem is to find a point T on the circle such that angle OTP is a right angle. Once T is found, the line PT is the tangent. The construction turns the search for a right angle into something a compass can do, using the fact that an inscribed angle on a diameter is a right angle.
The construction. Step 1: Draw segment OP with the straightedge. Step 2: Construct the midpoint M of OP by constructing the perpendicular bisector of OP (equal arcs from O and from P, above and below; the line through the two crossings meets OP at M). Step 3: With the compass point at M and radius MO = MP, draw a circle; it passes through both O and P. Step 4: This new circle meets the given circle in two points; call them T and T′. Step 5: Draw lines PT and PT′. Each is a tangent to the given circle from P.
Why the construction works. In the circle centered at M, the segment OP is a diameter, because M is its midpoint and the circle passes through O and P. The point T lies on that circle, so angle OTP is an inscribed angle on the diameter OP and therefore a right angle. But T also lies on the given circle, so OT is a radius of the given circle. A line through a point of a circle perpendicular to the radius at that point is the tangent there. Hence PT is tangent to the given circle at T. The same reasoning applies to T′. Note where each fact came from: "inscribed angle on a diameter is right" gave the right angle at T, and "tangent is perpendicular to radius" turned that right angle into tangency.
Why there are exactly two tangents. Two distinct circles meet in at most two points. The circle centered at M does meet the given circle in two points whenever P is outside the given circle: the M-circle passes through O, which is inside the given circle, and through P, which is outside it, so it has points on both sides of the given circle and must cross it, and a circle that crosses another does so at two points. If P were on the given circle there would be one tangent (the M-circle would be tangent to the given circle at P), and if P were inside there would be none, matching what we expect geometrically.
A worked model with coordinates. Let the given circle have center O(0, 0) and radius 6, and let P be (10, 0). The midpoint of OP is M(5, 0), and the construction circle has center M and radius 5: (x − 5)² + y² = 25. The given circle is x² + y² = 36. Subtracting the first equation from the second removes y²: x² − (x − 5)² = 36 − 25, so x² − (x² − 10x + 25) = 11, so 10x − 25 = 11 and x = 3.6. Then y² = 36 − 3.6² = 36 − 12.96 = 23.04, so y = ±4.8. The points of tangency are T(3.6, 4.8) and T′(3.6, −4.8). Check that T is on both circles: 3.6² + 4.8² = 12.96 + 23.04 = 36, and (3.6 − 5)² + 4.8² = 1.96 + 23.04 = 25. Check the right angle at T: the slope of OT is 4.8/3.6 = 4/3 and the slope of TP is (0 − 4.8)/(10 − 3.6) = −4.8/6.4 = −3/4, and (4/3)(−3/4) = −1, so OT ⟂ TP. Finally the tangent length: PT = √((10 − 3.6)² + 4.8²) = √(40.96 + 23.04) = √64 = 8, and indeed 6² + 8² = 36 + 64 = 100 = 10² = OP², the Pythagorean relation for the right triangle OTP.
Reading the result. The tangent length from P depends only on the distance OP and the radius r: PT² = OP² − r². For OP = 10 and r = 6, PT = 8. The construction did not need this formula, but the formula checks the construction, and it lets us compute tangent lengths without drawing. The two tangent segments PT and PT′ are congruent, as proved by hypotenuse-leg in the previous lesson; here both equal 8.
The misconception: draw the line from P to the "edge" of the circle. Students often place T where the line from P seems to graze the circle, or at the end of the radius perpendicular to OP. Neither is right. The radius perpendicular to OP ends at (0, 6) in the worked model, and the line from (10, 0) to (0, 6) has slope −3/5; the radius to (0, 6) is vertical, and the two are not perpendicular, so that line cuts the circle a second time. The true point of tangency T(3.6, 4.8) is closer to P than the top of the circle is, and it is found by the right-angle condition, not by eye.
A second misconception: the auxiliary circle is arbitrary. The circle centered at M with radius MO is not a decoration; it is the set of all points X for which angle OXP is a right angle (the locus of right angles on the diameter OP). Intersecting it with the given circle is exactly what selects the points where the radius and the line to P are perpendicular. Any other circle through O and P would not have OP as a diameter, and the angle at its intersection points would not be 90°.
Worked construction with proof: Given circle O and external point P: draw OP; construct its midpoint M; draw the circle with center M and radius MP; mark its intersections T and T′ with circle O; draw PT and PT′. Proof for PT: OP is a diameter of circle M, and T is on circle M, so angle OTP is inscribed on a diameter and equals 90°. T is on circle O, so OT is a radius of circle O. A line perpendicular to a radius at its endpoint on the circle is tangent to the circle, so PT is tangent at T.
Worked coordinates, second example: Circle O: center (0, 0), radius 5; P = (13, 0). M = (6.5, 0); circle M: (x − 6.5)² + y² = 42.25. Subtract from x² + y² = 25: x² − (x − 6.5)² = 25 − 42.25, so 13x − 42.25 = −17.25 and 13x = 25, x = 25/13 ≈ 1.923. Then y² = 25 − (25/13)² = 25 − 625/169 = (4225 − 625)/169 = 3600/169, so y = ±60/13 ≈ ±4.615. T = (25/13, 60/13). Tangent length: PT = √(13² − 5²) = √144 = 12. Check via coordinates: (13 − 25/13)² + (60/13)² = (144/13)² + (60/13)² = (20736 + 3600)/169 = 24336/169 = 144, so PT = 12.
Worked tangent-length problem: A point P is 15 cm from the center of a circle of radius 9 cm. The tangent segments from P have length √(15² − 9²) = √(225 − 81) = √144 = 12 cm. The two points of tangency, together with P and O, form a kite PTOT′ with right angles at T and T′; the angle between the two tangents at P satisfies cos(angle OPT) = 12/15 = 0.8, so angle OPT ≈ 36.87° and the full angle at P is about 73.74°.
Worked check of a wrong tangent: In the model with O(0, 0), r = 6 and P(10, 0), a student joins P to the top of the circle, (0, 6). The line has equation y = −(3/5)x + 6. Substituting into x² + y² = 36: x² + (9/25)x² − (36/5)x + 36 = 36, so (34/25)x² − (36/5)x = 0, giving x = 0 or x = 180/34 ≈ 5.29. The line meets the circle a second time at x ≈ 5.29, so it is a secant, not a tangent.
Construction record: Draw a circle of radius 4 cm and a point P 9 cm from its center. Carry out the five construction steps, leaving all arcs visible, and label O, P, M, T and T′. Measure PT and PT′ and compare them with the computed value √(9² − 4²) = √65 ≈ 8.06 cm. Write the two facts the proof uses beside the figure, in the order they are used.
Independent performance: Prepare a one-page guide titled "How to construct a tangent from an outside point, and why it works." It must contain the numbered steps, a labeled diagram, a proof that the constructed line is tangent, and a coordinate check using a circle and point of your own choosing in which you find the point of tangency by solving the two circle equations and verify that the radius and the tangent are perpendicular. Success criteria: the proof names the inscribed-angle-on-a-diameter fact and the tangent-perpendicular-to-radius fact explicitly; the coordinate check shows the subtraction of the two equations, both coordinates of T, and the slope product −1 (or, if the radius is vertical or horizontal, that the tangent is horizontal or vertical); and the tangent length is verified with PT² = OP² − r².
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
List the steps for constructing a tangent from an external point P to a circle with center O.
Draw OP. Construct the perpendicular bisector of OP to find its midpoint M. Draw the circle with center M and radius MO (it passes through O and P). Mark its intersections T and T′ with the given circle. Draw PT and PT′; each is tangent to the given circle.
Prove that the line PT produced by the construction is tangent to the given circle.
OP is a diameter of the circle centered at M, and T lies on that circle, so the inscribed angle OTP is a right angle. T also lies on the given circle, so OT is a radius. A line through a point of a circle that is perpendicular to the radius there is the tangent at that point, so PT is tangent at T.
The given circle is x² + y² = 36 and P = (10, 0). Find the coordinates of the points of tangency.
The construction circle has center (5, 0) and radius 5: (x − 5)² + y² = 25. Subtracting from x² + y² = 36 gives 10x − 25 = 11, so x = 3.6. Then y² = 36 − 12.96 = 23.04, y = ±4.8. The points are (3.6, 4.8) and (3.6, −4.8).
Verify that at T(3.6, 4.8) the radius OT and the line TP are perpendicular.
Slope of OT is 4.8/3.6 = 4/3. Slope of TP is (0 − 4.8)/(10 − 3.6) = −4.8/6.4 = −3/4. The product is (4/3)(−3/4) = −1, so the lines are perpendicular, which is the tangent condition.
Find the length of the tangent segment from a point 10 units from the center of a circle of radius 6, and explain why the formula applies.
The radius to the point of tangency is perpendicular to the tangent, so O, T and P form a right triangle with hypotenuse OP. PT² = 10² − 6² = 100 − 36 = 64, so PT = 8.
Explain why the construction circle must have OP as a diameter, not merely pass through O and P.
The right angle at T comes from the inscribed angle theorem: an inscribed angle is a right angle exactly when it intercepts a semicircle, that is, when the chord it stands on is a diameter. A circle through O and P with a different center would still pass through O and P, but OP would be an ordinary chord and the angle at the intersection points would not be 90°.
A student draws the segment from P(10, 0) to the top of the circle x² + y² = 36, the point (0, 6), and calls it a tangent. Show that it is not.
The line through (10, 0) and (0, 6) is y = −(3/5)x + 6. Substituting into the circle gives (34/25)x² − (36/5)x = 0, with solutions x = 0 and x = 180/34 ≈ 5.29. The line meets the circle at two points, so it is a secant. Equivalently, the radius to (0, 6) is vertical and the line is not horizontal, so they are not perpendicular.
How many tangents can be drawn to a circle from a point outside it, on it, and inside it? Explain using the construction.
Outside: two, because the construction circle crosses the given circle in two points. On the circle: one, the line perpendicular to the radius at that point (the construction circle is tangent to the given circle there). Inside: none, because every line through an interior point crosses the circle twice.