AMC 10 · 2019 · #23
Grade 8 geometry-2dProblem
Points A=(6,13) and B=(12,11) lie on circle ω in the plane. Suppose that the tangent lines to ω at A and B intersect at a point on the x-axis. What is the area of ω?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Points $A = (6, 13)$ and $B = (12, 11)$ lie on a circle $\omega$. The tangent lines to $\omega$ at $A$ and $B$ meet at a single point $P$ on the $x$-axis. Find the area of $\omega$.
Givens: $A = (6, 13)$ and $B = (12, 11)$ on circle $\omega$; Tangents to $\omega$ at $A$ and $B$ meet at some point $P = (t, 0)$ on the $x$-axis; Answer choices: (A) $\tfrac{83\pi}{8}$, (B) $\tfrac{21\pi}{2}$, (C) $\tfrac{85\pi}{8}$, (D) $\tfrac{43\pi}{4}$, (E) $\tfrac{87\pi}{8}$
Unknowns: The area of $\omega$, equivalently $\pi r^2$ where $r$ is the radius
Understand
Restated: Points $A = (6, 13)$ and $B = (12, 11)$ lie on a circle $\omega$. The tangent lines to $\omega$ at $A$ and $B$ meet at a single point $P$ on the $x$-axis. Find the area of $\omega$.
Givens: $A = (6, 13)$ and $B = (12, 11)$ on circle $\omega$; Tangents to $\omega$ at $A$ and $B$ meet at some point $P = (t, 0)$ on the $x$-axis; Answer choices: (A) $\tfrac{83\pi}{8}$, (B) $\tfrac{21\pi}{2}$, (C) $\tfrac{85\pi}{8}$, (D) $\tfrac{43\pi}{4}$, (E) $\tfrac{87\pi}{8}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #13 Convert to Algebra, #6 Guess and Check
Tool #1 (Diagram): the picture (circle, two tangent points, external point on the $x$-axis) clarifies that $|PA| = |PB|$ and that $\triangle OAP$ is right-angled at $A$. Tool #7 (Subproblems): split into (a) find $P$ on the $x$-axis from $|PA| = |PB|$, (b) compute $|PA|$, (c) use right-triangle/Ptolemy to relate $r$, $|PA|$, $|PO|$, (d) solve for $r^2$. Tool #13 (Algebra): each subproblem reduces to a short algebraic step (linear equation for $t$, distance squared, Pythagorean relation, Ptolemy).
Execute — Answer: C
7.G.B.4 Step 1 - Draw $A$, $B$, the circle $\omega$, the two tangent rays meeting at $P = (t, 0)$ on the $x$-axis, and the center $O$.
- Mark right angles at $A$ and $B$.
💡 Picture shows two right triangles $OAP, OBP$ with shared hypotenuse $OP$.
8.G.B.8 Step 2 - Find $t$.
- Equal tangent lengths: $|PA|^2 = |PB|^2$.
- Compute: $(t - 6)^2 + (0 - 13)^2 = (t - 12)^2 + (0 - 11)^2$.
- Expand: $t^2 - 12t + 36 + 169 = t^2 - 24t + 144 + 121$.
- Simplify: $-12t + 205 = -24t + 265$, so $12t = 60$, hence $t = 5$.
💡 Two tangents from one external point are equal — sets up a linear equation for $t$.
8.G.B.8 Step 3 - Compute the tangent length.
- $|PA|^2 = (6 - 5)^2 + (13 - 0)^2 = 1 + 169 = 170$.
- Also $|PB|^2 = (12 - 5)^2 + 11^2 = 49 + 121 = 170$ ✓.
💡 Distance squared by the Pythagorean theorem; equality double-checks $t = 5$.
8.G.B.8 Step 4 - Compute the chord length: $|AB|^2 = (12 - 6)^2 + (11 - 13)^2 = 36 + 4 = 40$, so $|AB| = 2\sqrt{10}$.
- Compute $|PM|$ where $M$ is the midpoint of $AB$.
- $M = (9, 12)$.
- $|PM|^2 = (9 - 5)^2 + (12 - 0)^2 = 16 + 144 = 160$, so $|PM| = 4\sqrt{10}$.
💡 We need $AB$ for Ptolemy and $PM$ to locate the center along line $PM$.
7.G.B.4 Step 5 - Use Ptolemy's relation on the cyclic quadrilateral $PAOB$.
- Since $\angle PAO = \angle PBO = 90^{\circ}$, $PAOB$ is inscribed in a circle of diameter $PO$.
- The diagonals are $PO$ and $AB$, and the sides are $PA = PB = \sqrt{170}$ and $OA = OB = r$.
- Ptolemy: $PA \cdot OB + PB \cdot OA = PO \cdot AB$, i.e.
- $2\sqrt{170} \cdot r = PO \cdot 2\sqrt{10}$, so $PO = \sqrt{17} \cdot r$.
💡 Ptolemy lets us turn the kite's right angles directly into a relation between $PO$ and $r$.
8.G.B.7 Step 6 - Apply Pythagoras to right triangle $OAP$: $OP^2 = OA^2 + AP^2$, i.e.
- $(r\sqrt{17})^2 = r^2 + 170$, so $17 r^2 = r^2 + 170$, hence $16 r^2 = 170$ and $r^2 = \tfrac{170}{16} = \tfrac{85}{8}$.
💡 The right triangle with the radius as one leg and the tangent length as the other pins down $r^2$.
7.G.B.4 Step 7 - Area of $\omega$ is $\pi r^2 = \tfrac{85 \pi}{8}$.
- The answer is (C).
💡 Match $\tfrac{85}{8}$ against the answer choices.
7.G.B.4 Draw $A$, $B$, the circle $\omega$, the two tangent rays meeting at $P = (t, 0)$ 8.G.B.8 Find $t$. Equal tangent lengths: $|PA|^2 = |PB|^2$. Compute: $(t - 6)^2 + (0 - 1 8.G.B.8 Compute the tangent length. $|PA|^2 = (6 - 5)^2 + (13 - 0)^2 = 1 + 169 = 170$. A 8.G.B.8 Compute the chord length: $|AB|^2 = (12 - 6)^2 + (11 - 13)^2 = 36 + 4 = 40$, so 7.G.B.4 Use Ptolemy's relation on the cyclic quadrilateral $PAOB$. Since $\angle PAO = \ 8.G.B.7 Apply Pythagoras to right triangle $OAP$: $OP^2 = OA^2 + AP^2$, i.e. $(r\sqrt{17 7.G.B.4 Area of $\omega$ is $\pi r^2 = \tfrac{85 \pi}{8}$. The answer is (C). Review
Reasonableness: Verify by independent coordinate computation. The center $O$ lies on the perpendicular bisector of $AB$, which has slope $-\tfrac{1}{3}$, so the bisector has slope $3$ and passes through $M = (9, 12)$: $y = 3x - 15$. The center also lies on line $PM$ from $P = (5, 0)$ through $M = (9, 12)$: that line is $y = 3(x - 5) = 3x - 15$ — same line ✓ (as it should be, since the locus of points equidistant from the two tangents through $A, B$ is exactly the line through $P$ and $M$). So $O$ is on $y = 3x - 15$, and additionally on the line through $A$ perpendicular to $PA$. $PA$ direction is $(1, 13)$, so the line through $A = (6, 13)$ perpendicular to $PA$ has direction $(-13, 1)$: parametrically $(6 - 13s, 13 + s)$. Plug into $y = 3x - 15$: $13 + s = 3(6 - 13s) - 15 = 3 - 39s$, so $40s = -10$, $s = -\tfrac{1}{4}$. Then $O = (6 + \tfrac{13}{4}, 13 - \tfrac{1}{4}) = (\tfrac{37}{4}, \tfrac{51}{4})$. Now $r^2 = (\tfrac{37}{4} - 6)^2 + (\tfrac{51}{4} - 13)^2 = (\tfrac{13}{4})^2 + (-\tfrac{1}{4})^2 = \tfrac{169 + 1}{16} = \tfrac{170}{16} = \tfrac{85}{8}$ ✓. Same value. Area = $\tfrac{85\pi}{8}$.
Alternative: Tool #13 (Algebra) without Ptolemy: parameterize the center $O = (a, b)$ with $|OA| = |OB|$ (perpendicular bisector of $AB$) and $|OA| = r$, $|OP| = \sqrt{r^2 + 170}$ from the right triangle $OAP$. Solve this system directly. The Ptolemy shortcut is faster, but both routes give $r^2 = \tfrac{85}{8}$.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Converting $r^2$ to the area $\pi r^2$ and recognizing the role of radius/tangent in the circle setup.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Right triangle $OAP$ with legs $OA = r$, $AP = \sqrt{170}$, hypotenuse $OP$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Distance-squared computations $|PA|^2$, $|PB|^2$, $|AB|^2$, $|PM|^2$, $r^2$.)
⭐ This AMC 10 problem only needs Grade 8 Pythagoras and the circle's basic 'radius ⟂ tangent' fact — solve the linear equation $|PA| = |PB|$ to find $P = (5, 0)$, get $|PA|^2 = 170$, then $16 r^2 = 170$ from a right-triangle/Ptolemy relation, so the area is $\tfrac{85 \pi}{8}$.
⭐ This AMC 10 problem only needs Grade 8 Pythagoras and the circle's basic 'radius ⟂ tangent' fact — solve the linear equation $|PA| = |PB|$ to find $P = (5, 0)$, get $|PA|^2 = 170$, then $16 r^2 = 170$ from a right-triangle/Ptolemy relation, so the area is $\tfrac{85 \pi}{8}$.
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