Analytic Geometry — Position of a Point Relative to a Circle
Analytic Geometry — Position of a Point Relative to a Circle. Practice questions to deepen understanding of the position of a point relative to a circle. Online math practice with full solutions and step-by-step explanations.
Position of a point relative to a circle practice — point inside / on / outside the circle, computing distance from the center, comparison to the radius. Analytic geometry.
📍 Position of a point:
How do you check whether a point \((x_0,y_0)\) lies on the circle \((x-a)^2+(y-b)^2=r^2\)?
Substitution test! 📍
| 📐 Method: Plug both coordinates into the canonical equation and compute the left-hand side: L = (x₀ − a)² + (y₀ − b)² Compare L with r²: • L = r² → on the circle • L < r² → strictly inside • L > r² → strictly outside 🎯 Why this works: The expression (x₀ − a)² + (y₀ − b)² is exactly the squared distance from the point to the centre. Comparing it to r² compares actual distance with the radius. ⚠️ Don''t do these instead: • Comparing the point to just the centre — irrelevant • Substituting only x₀ — you need both coordinates • Using the linear distance with √ — same idea but extra effort; squared distance saves a square root 💡 Remember: always compare to r² (the right-hand side), not to r. |
✓ On the circle:
Is the point \((4,1)\) on the circle \((x-1)^2+(y-1)^2=9\)?
Substituting and checking! ✓
| 📐 Substitute x = 4, y = 1: (4 − 1)² + (1 − 1)² = 3² + 0² = 9 + 0 = 9 Compare with r² = 9 → 9 = 9 ✓ The point is exactly on the circle. 🎯 Geometric check: The centre is (1, 1) and r = 3. Distance from (4, 1) to (1, 1) is √((4−1)² + (1−1)²) = √9 = 3 = radius. Confirms the point lies on the circle. 💡 Note: the y-coordinates of the centre and the point are identical (both 1), so the entire distance lies along the x-axis — easy to picture. |
🔵 Inside:
Where is the point \((2,3)\) relative to the circle \((x-2)^2+(y-2)^2=4\)?
The point is strictly inside! 🔵
| 📐 Substitute x = 2, y = 3: (2 − 2)² + (3 − 2)² = 0² + 1² = 1 Compare with r² = 4 → 1 < 4 → inside. 🎯 Distance check: Centre (2, 2), radius √4 = 2. Distance from (2, 3) to centre: √1 = 1, which is less than 2 (the radius), confirming "inside". 📊 The "inside" criterion: (x₀ − a)² + (y₀ − b)² < r² ⇔ point is strictly inside the disk bounded by the circle. 💡 Note: the centre itself satisfies LHS = 0 < r², so the centre is "inside" by this criterion — which matches intuition. |
🔴 Outside:
Where is the point \((7,3)\) relative to the circle \((x-2)^2+(y-3)^2=9\)?
The point is strictly outside! 🔴
| 📐 Substitute x = 7, y = 3: (7 − 2)² + (3 − 3)² = 5² + 0² = 25 Compare with r² = 9 → 25 > 9 → outside. 🎯 Distance check: Centre (2, 3), radius √9 = 3. Distance from (7, 3) to centre: √25 = 5, which is bigger than the radius 3, confirming "outside". 📊 The "outside" criterion: (x₀ − a)² + (y₀ − b)² > r² ⇔ point is strictly outside the disk. 💡 Geometric picture: the point and the centre share the same y-coordinate, so the distance is purely horizontal — 7 − 2 = 5 units, easily compared with the radius of 3. |
📏 Distance:
What is the distance from the point \((5,7)\) to the centre of the circle \((x-2)^2+(y-3)^2=25\)?
Computing the distance to the centre! 📏
| 📐 Identify centre and apply the distance formula: Centre: (2, 3). Point: (5, 7). d = √((5 − 2)² + (7 − 3)²) = √(9 + 16) = √25 = 5 🎯 Compare with the radius: r² = 25 → r = 5. So d = r — the point is exactly on the circle. 📊 The "3-4-5" triangle: The horizontal leg is 3 and the vertical leg is 4, giving the classic Pythagorean 3-4-5 right triangle. Hypotenuse = 5 = distance to the centre. 💡 Key takeaway: "distance to centre = radius" is the geometric definition of being on the circle. The substitution method (LHS = r²) is just a streamlined version of the same check. |
⚡ Quick check:
Where is the point \((3,3)\) relative to the circle \(x^2+y^2=25\)?
Quick substitution! ⚡
| 📐 Substitute x = 3, y = 3: 3² + 3² = 9 + 9 = 18 Compare with r² = 25 → 18 < 25 → inside. 🎯 Geometric picture: Origin-centred circle of radius 5. The point (3, 3) is at distance √18 ≈ 4.24 from the origin, which is less than the radius 5. 💡 Notice: for an origin-centred circle (a = b = 0), the test simplifies to "x₀² + y₀² vs r²", which is faster than the general (x₀ − a)² + (y₀ − b)² form. |
📐 On the axis:
Where is the point \((5,0)\) relative to the circle \(x^2+y^2=25\)?
Substitute and compare! 📐
| 📐 Substitute x = 5, y = 0: 5² + 0² = 25 + 0 = 25 Compare with r² = 25 → 25 = 25 → on the circle. 🎯 The point lives on two things at once: (5, 0) lies on the x-axis (y = 0) AND on the circle x² + y² = 25. It is one of two intersection points of the circle with the x-axis (the other being (−5, 0)). 💡 General insight: origin-centred circle x² + y² = r² always meets each axis at two diametrically-opposite points: (±r, 0) on the x-axis, (0, ±r) on the y-axis. |
🔄 Reverse:
If the point \((6,y)\) lies on the circle \((x-2)^2+(y-1)^2=25\), what is \(y\)?
Solving for the unknown coordinate! 🔄
| 📐 Substitute x = 6: (6 − 2)² + (y − 1)² = 25 16 + (y − 1)² = 25 (y − 1)² = 9 y − 1 = ±3 y = 1 + 3 = 4 or y = 1 − 3 = −2 🎯 Why two answers: A vertical line x = 6 meets the circle at two points (assuming the line lies within the circle''s range). These two points are mirror images of each other through the horizontal line y = 1 (which contains the centre). 📊 Verify: For (6, 4): (6 − 2)² + (4 − 1)² = 16 + 9 = 25 ✓ For (6, −2): (6 − 2)² + (−2 − 1)² = 16 + 9 = 25 ✓ 💡 Key: whenever you take a square root, allow both signs — you may get one, two, or zero solutions. |
⚠️ Mistake:
A student says: if \((x_0-a)^2+(y_0-b)^2=30\) and \(r=5\), the point is on the circle. Right?
r vs. r² confusion! ⚠️
| ❌ Why the student is wrong: The substitution gives the squared distance from the point to the centre, which must be compared with r² — not with r. r = 5 → r² = 25. 30 > 25 → the point is outside the circle, not on it. 📐 Right vs. wrong comparison: • Wrong: "30 ≠ 5, so not on the circle, but somewhere else" — yet they did call it "on" because the equation = 30 looked exotic. Confused logic. • Right: compare 30 with r² = 25. 30 > 25 → outside. 📊 Distance interpretation: Distance from the point to the centre is √30 ≈ 5.48, slightly larger than r = 5 → outside. 💡 Anti-trap rule: The right-hand side of the canonical equation is r squared, not r. Always square the radius before comparing. |
📚 Summary:
How do you determine the position of a point relative to a circle?
Point-position summary! 📚
| 🎯 The procedure: 1. Identify the canonical equation: (x − a)² + (y − b)² = r². 2. Substitute the point''s coordinates and compute the left-hand side L. 3. Compare L with r²:
⚠️ Top pitfalls: • Compare with r² (right-hand side), not r • Substitute both coordinates, never just one • Don''t skip the squaring step on r 💡 Why it works: the substituted value equals the squared distance from the point to the centre — directly the geometric relationship to the radius. |