Solutions — Circles

1. Key features from equations. Fluency

   • (a) x² + y² = 25: Centre (0, 0), radius 5. (r² = 25, so r = 5.)
   • (b) (x−3)² + (y−2)² = 16: Centre (3, 2), radius 4. (r² = 16, so r = 4.)
   • (c) (x+1)² + (y−4)² = 9: x+1 = x−(−1), so centre (−1, 4), radius 3.
   • (d) x² + (y+5)² = 1: Centre (0, −5), radius 1.

2. Write the equation. Fluency

   • (a) Centre (0,0), r=7: x² + y² = 49
   • (b) Centre (2,−3), r=5: (x − 2)² + (y + 3)² = 25
   • (c) Centre (−4,1), r=√3: r² = (√3)² = 3. (x + 4)² + (y − 1)² = 3
   • (d) Centre (0,6), r=2: x² + (y − 6)² = 4

3. Does the point lie on the circle? Fluency

   • (a) (3, 4): 3² + 4² = 9 + 16 = 25 ✓. Yes, (3, 4) lies on the circle.
   • (b) (0, 5): 0² + 5² = 25 ✓. Yes, (0, 5) lies on the circle.
   • (c) (1, 4): 1² + 4² = 1 + 16 = 17 ≠ 25. No, (1, 4) does not lie on the circle.
   • (d) (−4, 3): (−4)² + 3² = 16 + 9 = 25 ✓. Yes, (−4, 3) lies on the circle.

4. Domain and range. Fluency

   • (a) x² + y² = 36: r = 6. Domain: [−6, 6]. Range: [−6, 6].
   • (b) (x−1)² + (y+2)² = 25: Centre (1, −2), r = 5. Domain: [1−5, 1+5] = [−4, 6]. Range: [−2−5, −2+5] = [−7, 3].
   • (c) (x+3)² + y² = 4: Centre (−3, 0), r = 2. Domain: [−3−2, −3+2] = [−5, −1]. Range: [0−2, 0+2] = [−2, 2].
   • (d) x² + (y−3)² = 9: Centre (0, 3), r = 3. Domain: [0−3, 0+3] = [−3, 3]. Range: [3−3, 3+3] = [0, 6].

5. Completing the square. Understanding

   • (a) x²+y²−6x+2y−6=0: (x²−6x+9)+(y²+2y+1)=6+9+1=16. (x−3)²+(y+1)²=16. Centre (3,−1), radius 4.
   • (b) x²+y²+4x−8y+11=0: (x²+4x+4)+(y²−8y+16)=−11+4+16=9. (x+2)²+(y−4)²=9. Centre (−2,4), radius 3.
   • (c) x²+y²−2x−10y+22=0: (x²−2x+1)+(y²−10y+25)=−22+1+25=4. (x−1)²+(y−5)²=4. Centre (1,5), radius 2.
   • (d) x²+y²+8x+6y=0: (x²+8x+16)+(y²+6y+9)=16+9=25. (x+4)²+(y+3)²=25. Centre (−4,−3), radius 5.

6. Inside, on, or outside? Understanding

   • (a) (2, 8): (2−2)²+(8−3)²=0+25=25. Equal to 25 ⇒ on the circle.
   • (b) (5, 3): (5−2)²+(3−3)²=9+0=9<25. Less than 25 ⇒ inside the circle.
   • (c) (7, 7): (7−2)²+(7−3)²=25+16=41>25. Greater than 25 ⇒ outside the circle.
   • (d) (2, −2): (2−2)²+(−2−3)²=0+25=25. Equal to 25 ⇒ on the circle.

7. Equation from centre and point. Understanding

   • (a) Centre (0,0) through (5,12): r²=5²+12²=25+144=169. x²+y²=169
   • (b) Centre (3,−1) through (7,2): r²=(7−3)²+(2−(−1))²=16+9=25. (x−3)²+(y+1)²=25
   • (c) Centre (−2,4) through (1,4): r²=(1−(−2))²+(4−4)²=9+0=9. (x+2)²+(y−4)²=9
   • (d) Centre (0,−3) through (4,0): r²=(4−0)²+(0−(−3))²=16+9=25. x²+(y+3)²=25

8. Mobile tower coverage. Understanding

   • (a) Boundary equation: Centre (3, 2), r = 5. (x − 3)² + (y − 2)² = 25
   • (b) Town at (6, 6): (6−3)²+(6−2)²=9+16=25. The town is exactly on the boundary — just within coverage.
   • (c) House at (8, 3): (8−3)²+(3−2)²=25+1=26>25. The house is outside the coverage area — no coverage.
   • (d) Second tower at (−1, 2): Equation: (x+1)²+(y−2)²=25. Distance between centres: √[(3−(−1))²+(2−2)²]=√16=4 km. Sum of radii=5+5=10 km. Since 4<10, the coverage areas overlap.

9. Endpoints of a diameter. Problem Solving

   • (i) Midpoint of AB: Midpoint = ((1+7)/2, (7+1)/2) = (4, 4). This equals the centre of the circle. ✓
   • (ii) Check radius: |AB|²=(7−1)²+(1−7)²=36+36=72. |AB|=√72=6√2. Half of |AB| = 3√2. r²=(3√2)²=18, which matches the circle equation. ✓ So A and B are endpoints of a diameter.

10. Tangent to a circle. Problem Solving

    • (a) Gradient of radius to (3, 4): m = (4−0)/(3−0) = 4/3
    • (b) Tangent equation: Tangent is perpendicular ⇒ slope = −3/4. Using point (3,4): y−4=−3/4(x−3). Multiply by 4: 4y−16=−3x+9. 3x+4y=25