GCSE Maths: The equation of a circle and tangents (Higher)

Key words. The radius is the distance from the centre to the edge. A point is on a circle if its coordinates make the equation true.

Method. A circle with centre (0, 0) and radius r has equation x² + y² = r².

1. Remember the equation of a circle with its centre at the origin. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y make a right-angled triangle. Pythagoras gives x² + y² = (radius)².
   x² + y² = r²
2. Put in the radius. Here the radius is r = 3.
   r² = 3² = 9
3. Write the equation. Keep x² + y² on the left.
   x² + y² = 9

Check. Test the point (3, 0), which is on the circle: 3² + 0² = 9. ✓

Watch out. The right-hand side is r², not r. A circle of radius 5 has equation x² + y² = 25, not = 5.

Key words. The radius is the distance from the centre to the edge. A point is on a circle if its coordinates make the equation true.

Method. A circle with centre (0, 0) and radius r has equation x² + y² = r².

1. Remember the equation of a circle with its centre at the origin. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y make a right-angled triangle. Pythagoras gives x² + y² = (radius)².
   x² + y² = r²
2. Put in the radius. Here the radius is r = 6.
   r² = 6² = 36
3. Write the equation. Keep x² + y² on the left.
   x² + y² = 36

Check. Test the point (6, 0), which is on the circle: 6² + 0² = 36. ✓

Watch out. The right-hand side is r², not r. A circle of radius 5 has equation x² + y² = 25, not = 5.

Key words. The radius is the distance from the centre to the edge. A point is on a circle if its coordinates make the equation true.

Method. A circle with centre (0, 0) and radius r has equation x² + y² = r².

1. Remember the equation of a circle with its centre at the origin. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y make a right-angled triangle. Pythagoras gives x² + y² = (radius)².
   x² + y² = r²
2. Put in the radius. Here the radius is r = 10.
   r² = 10² = 100
3. Write the equation. Keep x² + y² on the left.
   x² + y² = 100

Check. Test the point (10, 0), which is on the circle: 10² + 0² = 100. ✓

Watch out. The right-hand side is r², not r. A circle of radius 5 has equation x² + y² = 25, not = 5.

Key words. The radius is the distance from the centre to the edge. A point is on a circle if its coordinates make the equation true.

Method. A circle with centre (0, 0) and radius r has equation x² + y² = r².

1. Remember the equation of a circle with its centre at the origin. Pick any point (x, y) on the circle. The radius to that point, the horizontal distance x and the vertical distance y make a right-angled triangle. Pythagoras gives x² + y² = (radius)².
   x² + y² = r²
2. Put in the radius. Here the radius is r = 2.5.
   r² = 2.5² = 6.25
3. Write the equation. Keep x² + y² on the left.
   x² + y² = 6.25

Check. Test the point (2.5, 0), which is on the circle: 2.5² + 0² = 6.25. ✓

Watch out. The right-hand side is r², not r. A circle of radius 5 has equation x² + y² = 25, not = 5.

Method. Read off r² from the equation, then take the square root to find the radius.

1. Compare with the standard form. The equation of a circle with centre the origin is x² + y² = r². The number on the right is the radius squared.
   r² = 49
2. Take the square root. The radius is a length, so it is positive.
   r = √49 = 7

Check. Square the answer: (7)² = 49. ✓

Watch out. The number on the right is r². Do not give 36 as the radius of x² + y² = 36; the radius is 6.

Method. Read off r² from the equation, then take the square root to find the radius.

1. Compare with the standard form. The equation of a circle with centre the origin is x² + y² = r². The number on the right is the radius squared.
   r² = 100
2. Take the square root. The radius is a length, so it is positive.
   r = √100 = 10

Check. Square the answer: (10)² = 100. ✓

Watch out. The number on the right is r². Do not give 36 as the radius of x² + y² = 36; the radius is 6.

Method. Read off r² from the equation, then take the square root to find the radius.

1. Compare with the standard form. The equation of a circle with centre the origin is x² + y² = r². The number on the right is the radius squared.
   r² = 2.25
2. Take the square root. The radius is a length, so it is positive.
   r = √2.25 = 32

Check. Square the answer: (32)² = 2.25. ✓

Watch out. The number on the right is r². Do not give 36 as the radius of x² + y² = 36; the radius is 6.

Method. Read off r² from the equation, then take the square root to find the radius.

1. Compare with the standard form. The equation of a circle with centre the origin is x² + y² = r². The number on the right is the radius squared.
   r² = 20
2. Take the square root. The radius is a length, so it is positive.
   r = √20 = 2√5
3. Check the surd is simplified. Look for a square factor.
   2√5 ≈ 4.47

Check. Square the answer: (2√5)² = 20. ✓

Watch out. The number on the right is r². Do not give 36 as the radius of x² + y² = 36; the radius is 6.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (3)² + (4)² = 9 + 16 = 25
2. Compare with r². The circle has r² = 25. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   25 = 25
3. Write the conclusion.
   The point (3, 4) is on the circle.

Check. The distance from the origin is √25 ≈ 5.00 and the radius is 5.00: equal. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (−5)² + (0)² = 25 + 0 = 25
2. Compare with r². The circle has r² = 25. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   25 = 25
3. Write the conclusion.
   The point (−5, 0) is on the circle.

Check. The distance from the origin is √25 ≈ 5.00 and the radius is 5.00: equal. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (2)² + (3)² = 4 + 9 = 13
2. Compare with r². The circle has r² = 25. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   13 < 25
3. Write the conclusion.
   The point (2, 3) is inside the circle.

Check. The distance from the origin is √13 ≈ 3.61 and the radius is 5.00: smaller, so inside. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (4)² + (4)² = 16 + 16 = 32
2. Compare with r². The circle has r² = 25. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   32 > 25
3. Write the conclusion.
   The point (4, 4) is outside the circle.

Check. The distance from the origin is √32 ≈ 5.66 and the radius is 5.00: bigger, so outside. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. Substitute the known coordinate into x² + y² = r², then solve for the other. There are usually two answers.

1. Substitute the known coordinate into the circle equation. The point is on the circle, so its coordinates satisfy the equation.
   x = 3:   (3)² + y² = 25
2. Work out the square and rearrange for y². Subtract the known square from both sides.
   9 + y² = 25
   y² = 25 − 9 = 16
3. Square root both sides. Remember there are two roots: positive and negative. This is because there are two points on the circle with this x-coordinate, one on each side.
   y = ±√16 = ±4

Check. Check one: (3)² + (4)² = 9 + 16 = 25. ✓

Watch out. A square root has two answers, positive and negative.

Method. Substitute the known coordinate into x² + y² = r², then solve for the other. There are usually two answers.

1. Substitute the known coordinate into the circle equation. The point is on the circle, so its coordinates satisfy the equation.
   x = −4:   ((−4))² + y² = 25
2. Work out the square and rearrange for y². Subtract the known square from both sides.
   16 + y² = 25
   y² = 25 − 16 = 9
3. Square root both sides. Remember there are two roots: positive and negative. This is because there are two points on the circle with this x-coordinate, one on each side.
   y = ±√9 = ±3

Check. Check one: ((−4))² + (3)² = 16 + 9 = 25. ✓

Watch out. A square root has two answers, positive and negative.

Method. Substitute the known coordinate into x² + y² = r², then solve for the other. There are usually two answers.

1. Substitute the known coordinate into the circle equation. The point is on the circle, so its coordinates satisfy the equation.
   x = 2:   (2)² + y² = 13
2. Work out the square and rearrange for y². Subtract the known square from both sides.
   4 + y² = 13
   y² = 13 − 4 = 9
3. Square root both sides. Remember there are two roots: positive and negative. This is because there are two points on the circle with this x-coordinate, one on each side.
   y = ±√9 = ±3

Check. Check one: (2)² + (3)² = 4 + 9 = 13. ✓

Watch out. A square root has two answers, positive and negative.

Method. Substitute the known coordinate into x² + y² = r², then solve for the other. There are usually two answers.

1. Substitute the known coordinate into the circle equation. The point is on the circle, so its coordinates satisfy the equation.
   x = 2:   (2)² + y² = 20
2. Work out the square and rearrange for y². Subtract the known square from both sides.
   4 + y² = 20
   y² = 20 − 4 = 16
3. Square root both sides. Remember there are two roots: positive and negative. This is because there are two points on the circle with this x-coordinate, one on each side.
   y = ±√16 = ±4

Check. Check one: (2)² + (4)² = 4 + 16 = 20. ✓

Watch out. A square root has two answers, positive and negative.

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   (3)² + (4)² = 9 + 16 = 25 ✓
2. Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
   gradient of radius = (4 − 0) ÷ (3 − 0) = 43
3. Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
   gradient of tangent = −1 ÷ 43 = −34
4. Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
   y − 4 = −34(x − 3)
5. Expand and rearrange to y = mx + c. Multiply out the bracket: −34 × 3 = −94. Then add 4 to both sides.
   y = −34x + 94 + 4
   y = −3x4 + 254
6. (Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
   3x + 4y = 25

Check. Put the point back in: y = −34 × 3 + 254 = 4, and the point has y = 4. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   ((−4))² + (3)² = 16 + 9 = 25 ✓
2. Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
   gradient of radius = (3 − 0) ÷ ((−4) − 0) = −34
3. Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
   gradient of tangent = −1 ÷ −34 = 43
4. Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
   y − 3 = 43(x − (−4))
5. Expand and rearrange to y = mx + c. Multiply out the bracket: 43 × (−4) = −163. Then add 3 to both sides.
   y = 43x + 163 + 3
   y = 4x3 + 253
6. (Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
   −4x + 3y = 25

Check. Put the point back in: y = 43 × (−4) + 253 = 3, and the point has y = 3. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   (5)² + (12)² = 25 + 144 = 169 ✓
2. Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
   gradient of radius = (12 − 0) ÷ (5 − 0) = 125
3. Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
   gradient of tangent = −1 ÷ 125 = −512
4. Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
   y − 12 = −512(x − 5)
5. Expand and rearrange to y = mx + c. Multiply out the bracket: −512 × 5 = −2512. Then add 12 to both sides.
   y = −512x + 2512 + 12
   y = −5x12 + 16912
6. (Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
   5x + 12y = 169

Check. Put the point back in: y = −512 × 5 + 16912 = 12, and the point has y = 12. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   ((−6))² + ((−8))² = 36 + 64 = 100 ✓
2. Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
   gradient of radius = ((−8) − 0) ÷ ((−6) − 0) = 43
3. Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
   gradient of tangent = −1 ÷ 43 = −34
4. Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
   y − (−8) = −34(x − (−6))
5. Expand and rearrange to y = mx + c. Multiply out the bracket: −34 × (−6) = 92. Then add (−8) to both sides.
   y = −34x − 92 + (−8)
   y = −3x4 − 252
6. (Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
   −6x − 8y = 100

Check. Put the point back in: y = −34 × (−6) + −252 = −8, and the point has y = −8. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Method. A tangent is perpendicular to the radius at the point of contact.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   (7)² + (0)² = 49 + 0 = 49 ✓
2. Think about the picture. The radius to (7, 0) lies along the x-axis (horizontal). A tangent is perpendicular to the radius, so it is vertical.
   A vertical line through x = 7
3. Write the equation. Every point on a vertical line has the same x-coordinate.
   x = 7

Check. The line x = 7 touches the circle only at (7, 0). ✓

Method. A tangent is perpendicular to the radius at the point of contact.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   (0)² + ((−6))² = 0 + 36 = 36 ✓
2. Think about the picture. The radius to (0, −6) lies along the y-axis (vertical). A tangent is perpendicular to the radius, so it is horizontal.
   A horizontal line through y = −6
3. Write the equation. Every point on a horizontal line has the same y-coordinate.
   y = −6

Check. The line y = −6 touches the circle only at (0, −6). ✓

Method. A tangent is perpendicular to the radius at the point of contact.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   ((−8))² + (0)² = 64 + 0 = 64 ✓
2. Think about the picture. The radius to (−8, 0) lies along the x-axis (horizontal). A tangent is perpendicular to the radius, so it is vertical.
   A vertical line through x = −8
3. Write the equation. Every point on a vertical line has the same x-coordinate.
   x = −8

Check. The line x = −8 touches the circle only at (−8, 0). ✓

Method. Write the tangent as ax + by = r², find its intercepts with the axes by setting x = 0 and y = 0, then use them as the base and height of a right-angled triangle.

1. Use the neat form of the tangent. At the point (a, b) on x² + y² = r², the tangent is ax + by = r² (this comes from the gradient method above).
   3x + 4y = 25
2. Find where it crosses the x-axis (put y = 0). On the x-axis the y-coordinate is 0.
   3x = 25 ⇒ x = 253
3. Find where it crosses the y-axis (put x = 0). On the y-axis the x-coordinate is 0.
   4y = 25 ⇒ y = 254
4. Find the area of the triangle with the axes. The triangle has base along the x-axis and height up the y-axis. Area = ½ × base × height.
   area = ½ × 253 × 254 = 62524

Check. The intercepts (253, 0) and (0, 254) both satisfy 3x + 4y = 25. ✓

Method. Write the tangent as ax + by = r², find its intercepts with the axes by setting x = 0 and y = 0, then use them as the base and height of a right-angled triangle.

1. Use the neat form of the tangent. At the point (a, b) on x² + y² = r², the tangent is ax + by = r² (this comes from the gradient method above).
   6x + 8y = 100
2. Find where it crosses the x-axis (put y = 0). On the x-axis the y-coordinate is 0.
   6x = 100 ⇒ x = 503
3. Find where it crosses the y-axis (put x = 0). On the y-axis the x-coordinate is 0.
   8y = 100 ⇒ y = 252
4. Find the area of the triangle with the axes. The triangle has base along the x-axis and height up the y-axis. Area = ½ × base × height.
   area = ½ × 503 × 252 = 6256

Check. The intercepts (503, 0) and (0, 252) both satisfy 6x + 8y = 100. ✓

Method. Substitute the equation of the line into the circle equation. A tangent touches at one point only, so the resulting quadratic has exactly one solution (a repeated root).

1. Rearrange the line to make y the subject. We will substitute this into the circle equation.
   3x + 4y = 25 ⇒ y = −3x + 254
2. Substitute into the circle equation. Where the line meets the circle, both equations are true at once. Replace y in x² + y² = 25.
   x² + (−3x + 254)² = 25
3. Expand, clear the fractions and simplify to a quadratic equal to 0. Multiply every term by 16 to clear the fractions, then move everything to one side.
   x² − 6x + 9 = 0
4. Factorise. Look for a perfect square.
   (x − 3)² = 0
5. Count the solutions. A perfect square gives only one value of x, so the line meets the circle at exactly one point. That is what a tangent does.
   x = 3 only. (Discriminant b² − 4ac = 0.)

Check. The point of contact is (3, 4); it satisfies both equations. ✓

Watch out. If the discriminant were positive the line would cut the circle twice (a secant); if negative it would miss the circle.

Method. Substitute the equation of the line into the circle equation. A tangent touches at one point only, so the resulting quadratic has exactly one solution (a repeated root).

1. Rearrange the line to make y the subject. We will substitute this into the circle equation.
   4x + 3y = 25 ⇒ y = −4x + 253
2. Substitute into the circle equation. Where the line meets the circle, both equations are true at once. Replace y in x² + y² = 25.
   x² + (−4x + 253)² = 25
3. Expand, clear the fractions and simplify to a quadratic equal to 0. Multiply every term by 9 to clear the fractions, then move everything to one side.
   x² − 8x + 16 = 0
4. Factorise. Look for a perfect square.
   (x − 4)² = 0
5. Count the solutions. A perfect square gives only one value of x, so the line meets the circle at exactly one point. That is what a tangent does.
   x = 4 only. (Discriminant b² − 4ac = 0.)

Check. The point of contact is (4, 3); it satisfies both equations. ✓

Watch out. If the discriminant were positive the line would cut the circle twice (a secant); if negative it would miss the circle.

Method. Substitute the equation of the line into the circle equation. A tangent touches at one point only, so the resulting quadratic has exactly one solution (a repeated root).

1. Rearrange the line to make y the subject. We will substitute this into the circle equation.
   5x + 12y = 169 ⇒ y = −5x + 16912
2. Substitute into the circle equation. Where the line meets the circle, both equations are true at once. Replace y in x² + y² = 169.
   x² + (−5x + 16912)² = 169
3. Expand, clear the fractions and simplify to a quadratic equal to 0. Multiply every term by 144 to clear the fractions, then move everything to one side.
   x² − 10x + 25 = 0
4. Factorise. Look for a perfect square.
   (x − 5)² = 0
5. Count the solutions. A perfect square gives only one value of x, so the line meets the circle at exactly one point. That is what a tangent does.
   x = 5 only. (Discriminant b² − 4ac = 0.)

Check. The point of contact is (5, 12); it satisfies both equations. ✓

Watch out. If the discriminant were positive the line would cut the circle twice (a secant); if negative it would miss the circle.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (5)² + (6)² = 25 + 36 = 61
2. Compare with r². The circle has r² = 64. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   61 < 64
3. Write the conclusion.
   The point (5, 6) is inside the circle.

Check. The distance from the origin is √61 ≈ 7.81 and the radius is 8.00: smaller, so inside. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Key words. A tangent is a straight line that touches a circle at exactly one point. It meets the radius at that point at a right angle.

Method. The tangent is perpendicular to the radius. Find the gradient of the radius, flip it and change the sign, then use the point to find the equation.

1. Check the point is on the circle. The tangent touches the circle at the point, so the point must satisfy the circle equation.
   (4.8)² + (6.4)² = 23.04 + 40.96 = 64 ✓
2. Find the gradient of the radius. The radius joins the centre (0, 0) to the point. Gradient = change in y ÷ change in x.
   gradient of radius = (6.4 − 0) ÷ (4.8 − 0) = 43
3. Find the gradient of the tangent. A tangent is perpendicular to the radius at the point of contact (Chapter 26). For perpendicular lines the gradients multiply to give −1: flip the fraction and change the sign.
   gradient of tangent = −1 ÷ 43 = −34
4. Use y − y₁ = m(x − x₁) with the point. The tangent has gradient m and passes through (a, b).
   y − 6.4 = −34(x − 4.8)
5. Expand and rearrange to y = mx + c. Multiply out the bracket: −34 × 4.8 = −185. Then add 6.4 to both sides.
   y = −34x + 185 + 6.4
   y = −3x4 + 10
6. (Optional) A neater form. Multiplying through by the bottom of the fraction removes the fractions. The result always has this pattern: ax + by = r².
   4.8x + 6.4y = 64

Check. Put the point back in: y = −34 × 4.8 + 10 = 325, and the point has y = 6.4. ✓

Watch out. The tangent gradient is the negative reciprocal of the radius gradient. A common slip is to flip the fraction but forget the sign change.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (6)² + (8)² = 36 + 64 = 100
2. Compare with r². The circle has r² = 100. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   100 = 100
3. Write the conclusion.
   The point (6, 8) is on the circle.

Check. The distance from the origin is √100 ≈ 10.00 and the radius is 10.00: equal. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (7)² + (7)² = 49 + 49 = 98
2. Compare with r². The circle has r² = 100. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   98 < 100
3. Write the conclusion.
   The point (7, 7) is inside the circle.

Check. The distance from the origin is √98 ≈ 9.90 and the radius is 10.00: smaller, so inside. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.

Method. Work out x² + y² for the point and compare it with r².

1. Substitute the coordinates into x² + y². If the point is on the circle, the left side will equal the right side.
   (−9)² + (4)² = 81 + 16 = 97
2. Compare with r². The circle has r² = 100. If the answer is equal, the point is on the circle; if it is smaller, the point is inside (closer to the centre); if bigger, outside.
   97 < 100
3. Write the conclusion.
   The point (−9, 4) is inside the circle.

Check. The distance from the origin is √97 ≈ 9.85 and the radius is 10.00: smaller, so inside. ✓

Watch out. Square the negative coordinates carefully: (−3)² = 9, not −9.