Analytic Geometry — Intersection of a Line with a Circle

Analytic Geometry — Intersection of a Line with a Circle. Practice questions to deepen understanding of the intersection of a line with a circle. Online math practice with full solutions and step-by-step explanations.

Intersection of a line with a circle practice — number of intersection points (0/1/2), substitution and discriminant, tangent as a special case. Analytic geometry.

10 questions

Question 1

10.00 pts

📐 Line-circle intersection:

How do you find the intersection points of a line \(y=mx+n\) with a circle?

Substitute \(y=mx+n\) into the circle equation and solve a quadratic equation

Substitute \(x=0\)

Compute only the distance

Compare the coefficients

Explanation:

💡 Detailed Explanation:

The substitution method! 📐

📐 Step-by-step procedure:
1️⃣ Take y from the line equation: y = mx + n
2️⃣ Substitute into the circle: (x − a)² + (mx + n − b)² = r²
3️⃣ Expand and collect terms in x — you get a quadratic Ax² + Bx + C = 0
4️⃣ Solve the quadratic for x
5️⃣ Plug each x back into y = mx + n to find y

🎯 Why a quadratic:
Both terms in the circle equation are squared, and y is replaced by a linear expression in x — squaring makes the equation degree 2 in x. The roots of this quadratic are the x-coordinates of the intersection points.

📊 Number of intersections:
From the discriminant Δ:
• Δ > 0 → 2 distinct intersection points (line crosses the circle)
• Δ = 0 → 1 point (line is tangent to the circle)
• Δ < 0 → 0 real points (line misses the circle)

💡 Vertical lines: if the line is x = k, you can''t solve for y from a slope-intercept form — substitute x = k directly into the circle equation and solve a quadratic in y.

Question 2

10.00 pts

🔢 Discriminant:

If the discriminant \(\Delta > 0\), how many intersection points are there?

\(2\) intersection points — the line crosses the circle

\(1\) point — tangent

\(0\) points

\(3\) points

Explanation:

💡 Detailed Explanation:

Discriminant tells the geometry! 🔢

📐 Why two intersections:
The substitution turns line ∩ circle into a quadratic Ax² + Bx + C = 0. A quadratic has Δ = B² − 4AC. When Δ > 0, the quadratic has two distinct real roots → two distinct x-values → two intersection points.

🎯 Geometric picture:
The line enters the circle at one point and exits at another — like a chord cutting through a disk.

💡 Geometric verification: Δ > 0 ↔ distance from centre to line < r ↔ line passes through the disk''s interior.

Question 3

10.00 pts

📏 Tangent:

When is a line tangent to a circle?

When \(\Delta = 0\) — discriminant is zero

When \(\Delta > 0\)

When \(\Delta < 0\)

Always

Explanation:

💡 Detailed Explanation:

Tangency means a single intersection! 📏

📐 Algebraic condition:
The substitution gives a quadratic in x. A tangent line meets the circle at exactly one point — that''s a quadratic with a single (repeated) root, so Δ = 0.

🎯 Equivalent geometric condition:
A tangent line is perpendicular to the radius at the point of tangency, and the distance from the centre to the line equals the radius:
distance(centre, line) = r

📊 Three formulations of "tangent":
1. The line meets the circle at a single point.
2. The discriminant of the substitution-quadratic is zero.
3. The perpendicular distance from the centre to the line equals the radius.

All three are equivalent — different lenses on the same idea.

Question 4

10.00 pts

❌ No intersection:

When does a line miss a circle entirely?

When \(\Delta < 0\) — discriminant is negative

When \(\Delta = 0\)

When \(\Delta > 0\)

Never

Explanation:

💡 Detailed Explanation:

No real intersection! ❌

📐 Algebraic test:
If the substitution-quadratic has Δ < 0, its roots are complex (not real). Geometrically there are no real intersection points — the line completely misses the circle.

🎯 Geometric equivalent:
distance(centre, line) > r — the closest point on the line to the centre is still farther than the radius. The line lies entirely outside the disk.

📊 Mini-example:
Circle x² + y² = 1, line y = 5.
Substitution: x² + 25 = 1 → x² = −24. No real solutions → no intersection. Geometrically y = 5 is far above the unit circle.

Question 5

10.00 pts

🔢 Compute:

Find the intersection points of \(y=2x\) with \(x^2+y^2=20\).

\((2,4)\) and \((-2,-4)\)

\((4,8)\) and \((-4,-8)\)

\((1,2)\) and \((-1,-2)\)

No intersection

Explanation:

💡 Detailed Explanation:

Substitution and quadratic! 🔢

📐 Step-by-step:
1️⃣ Substitute y = 2x into x² + y² = 20:
x² + (2x)² = 20
x² + 4x² = 20
5x² = 20 → x² = 4 → x = ±2

2️⃣ Find corresponding y:
For x = 2: y = 4 → (2, 4)
For x = −2: y = −4 → (−2, −4)

✅ Verify: 2² + 4² = 20 ✓ and (−2)² + (−4)² = 20 ✓

💡 Symmetry: both line and circle pass through the origin and are symmetric about it → the two intersection points are diametrically opposite (mirror images through the origin).

Question 6

10.00 pts

📏 Vertical line:

Find the intersection of \(x=3\) with \((x-1)^2+(y-2)^2=20\).

\((3,6)\) and \((3,-2)\)

\((3,2)\)

\((1,2)\)

No intersection

Explanation:

💡 Detailed Explanation:

Vertical line — substitute and solve in y! 📏

📐 Substitute x = 3:
(3 − 1)² + (y − 2)² = 20
4 + (y − 2)² = 20
(y − 2)² = 16
y − 2 = ±4
y = 6 or y = −2

Intersection points: (3, 6) and (3, −2).

📊 Geometric check: centre (1, 2), radius √20 ≈ 4.47. The line x = 3 is 2 units to the right of the centre, well within the radius — so two intersections expected.

💡 Symmetry: the two intersection points are vertically symmetric about y = 2 (the y-coordinate of the centre).

Question 7

10.00 pts

➡️ Horizontal line:

Does \(y=7\) intersect \(x^2+y^2=25\)?

No — because \(7 > 5\), the line is above the circle

Yes, at one point

Yes, at two points

Depends on the centre

Explanation:

💡 Detailed Explanation:

The line is too far away! ➡️

📐 Quick check:
Circle x² + y² = 25 has centre (0, 0) and radius 5. The horizontal line y = 7 is at distance 7 from the centre — greater than 5 → no intersection.

🎯 Algebraic confirmation:
Substitute y = 7: x² + 49 = 25 → x² = −24 → no real solution.

💡 The boundary cases:
• y = 5 → tangent at (0, 5)
• |y| < 5 → secant (two intersections)
• |y| > 5 → no intersections

Question 8

10.00 pts

🎯 Shifted circle:

Find the intersection of \(y=x\) with \((x-2)^2+(y-2)^2=8\).

\((0,0)\) and \((4,4)\)

\((2,2)\)

\((1,1)\) and \((3,3)\)

No intersection

Explanation:

💡 Detailed Explanation:

Substitution gives a quadratic! 🎯

📐 Substitute y = x:
(x − 2)² + (x − 2)² = 8
2(x − 2)² = 8 → (x − 2)² = 4
x − 2 = ±2 → x = 0 or x = 4

Points: (0, 0) and (4, 4).

📊 Geometric note: the line y = x passes through the centre (2, 2). A line through the centre always meets the circle at two diametrically-opposite points → (0, 0) and (4, 4) are endpoints of a diameter.

💡 Cross-check: 2r = 2√8 = 4√2. Distance from (0, 0) to (4, 4) = √32 = 4√2 ✓

Question 9

10.00 pts

⚠️ Mistake:

A student found \(\Delta = 4\) and said: there are \(4\) intersection points. Right?

No! \(\Delta > 0\) means \(2\) points, not \(4\)

Yes, correct

Depends on the circle

Need more information

Explanation:

💡 Detailed Explanation:

The discriminant value vs. the number of roots! ⚠️

❌ Why "4" is wrong:
The student confused the value of the discriminant with the number of intersection points. Δ is just a number — its sign tells you the type of intersection, not its count.

📐 Correct interpretation:
Δ = 4 > 0 → 2 distinct real roots → 2 intersection points.
Whether Δ is 0.001, 4, or 1,000,000, as long as it is positive, the answer is 2.

💡 Geometric upper bound: a line and a circle in the plane can intersect in at most 2 points. There is no scenario where they meet in 3 or 4 points.

Question 10

10.00 pts

📚 Summary:

What is the correct order to find the intersection of a line and a circle?

Substitute → quadratic equation → discriminant → solve

Discriminant → substitute → solve

Solve → substitute → discriminant

Just substitute

Explanation:

💡 Detailed Explanation:

The standard procedure! 📚

🎯 Four-step recipe:

1️⃣ Substitute y = mx + n into the circle equation
2️⃣ Quadratic: expand and collect to Ax² + Bx + C = 0
3️⃣ Discriminant: compute Δ to learn how many solutions exist
4️⃣ Solve: use the quadratic formula for x, then back-substitute for y

⚠️ Common pitfalls:
• Forgetting to back-substitute for y after solving for x
• Computing Δ before forming the standard-form quadratic
• Skipping verification (substituting the result back into both equations)

💡 Special cases:
• Vertical line x = k → substitute, get a quadratic in y
• Line through the centre → 2 intersections automatically; endpoints are diametrically opposite