▶️ Answer/Explanation
Solution
Part (a): Show centre lies on \( y = \frac{1}{2}x – 4 \)
Centre C(h, k), radius 10. Distance CA = CB = 10.
CA: \( (h – 4)^2 + (k – 3)^2 = 100 \) (1)
CB: \( (h – 8)^2 + (k + 5)^2 = 100 \) (2)
Expand (1): \( h^2 – 8h + 16 + k^2 – 6k + 9 = 100 \)
\( h^2 + k^2 – 8h – 6k – 75 = 0 \) (3)
Expand (2): \( h^2 – 16h + 64 + k^2 + 10k + 25 = 100 \)
\( h^2 + k^2 – 16h + 10k – 11 = 0 \) (4)
Subtract (3) from (4): \( -16h + 10k – 11 + 8h + 6k + 75 = 0 \)
\( -8h + 16k + 64 = 0 \)
\( k = \frac{1}{2}h – 4 \)
Answer: Centre lies on \( y = \frac{1}{2}x – 4 \).
Part (b): Equations of the circle
Substitute \( k = \frac{1}{2}h – 4 \) into (3):
\( h^2 + \left(\frac{1}{2}h – 4\right)^2 – 8h – 6\left(\frac{1}{2}h – 4\right) – 75 = 0 \)
\( h^2 + \frac{1}{4}h^2 – 4h + 16 – 8h – 3h + 24 – 75 = 0 \)
\( \frac{5}{4}h^2 – 15h – 35 = 0 \)
\( h^2 – 12h – 28 = 0 \)
\( h = \frac{12 \pm \sqrt{256}}{2} = 14, -2 \)
For \( h = 14 \): \( k = \frac{14}{2} – 4 = 3 \), centre (14, 3).
For \( h = -2 \): \( k = \frac{-2}{2} – 4 = -5 \), centre (-2, -5).
Equations:
\( (x – 14)^2 + (y – 3)^2 = 100 \)
\( (x + 2)^2 + (y + 5)^2 = 100 \)
Answer: \( (x – 14)^2 + (y – 3)^2 = 100 \), \( (x + 2)^2 + (y + 5)^2 = 100 \)
▶️ Answer/Explanation
Solution
Part (a): Find \( k \), \( p \), and other intersection point
Line at \( \left( \frac{1}{2}, \frac{5}{2} \right) \): \( \frac{5}{2} = \frac{k}{2} + p \) (1)
Curve at \( \left( \frac{1}{2}, \frac{5}{2} \right) \): \( \frac{5}{2} = \frac{k}{8} – k + 2 \)
Simplify: \( \frac{1}{2} = \frac{k – 8k}{8} \), \( \frac{1}{2} = -\frac{7k}{8} \), \( k = \frac{4}{7} \)
From (1): \( \frac{5}{2} = \frac{4}{14} + p \), \( p = \frac{5}{2} – \frac{2}{7} = \frac{31}{14} \)
Equate: \( \frac{2}{7}x^2 – \frac{8}{7}x + 2 = \frac{4}{7}x + \frac{31}{14} \)
Multiply by 14: \( 4x^2 – 16x + 28 = 8x + 31 \)
\( 4x^2 – 24x – 3 = 0 \)
Solve: \( x = \frac{24 \pm \sqrt{576 + 48}}{8} = \frac{24 \pm 4\sqrt{39}}{8} = \frac{6 \pm \sqrt{39}}{2} \)
Other root: \( x = \frac{6 + \sqrt{39}}{2} \)
\( y = \frac{4}{7} \cdot \frac{6 + \sqrt{39}}{2} + \frac{31}{14} = \frac{12 + 2\sqrt{39} + 31}{14} = \frac{43 + 2\sqrt{39}}{14} \)
Answer: \( k = \frac{4}{7} \), \( p = \frac{31}{14} \), other point: \( \left( \frac{6 + \sqrt{39}}{2}, \frac{43 + 2\sqrt{39}}{14} \right) \)
Part (b): Possible values of \( p \) for no intersection
Equate: \( \frac{1}{2}kx^2 – 2kx + 2 = kx + p \)
\( \frac{1}{2}kx^2 – 3kx + 2 – p = 0 \)
Discriminant: \( \Delta = 9k^2 – 2k(2 – p) = 9k^2 – 4k + 2kp \)
For no roots: \( 9k^2 – 4k + 2kp < 0 \)
\( p < \frac{-9k + 4}{2} \)
Answer: \( p < \frac{4 – 9k}{2} \)
▶️ Answer/Explanation
Solution
Part (a): Express circle equation in standard form
Circle equation: \( x^2 + y^2 + px + 2y + q = 0 \)
Complete the square for \( x \):
\( x^2 + px = \left( x + \frac{p}{2} \right)^2 – \left( \frac{p}{2} \right)^2 \)
For \( y \):
\( y^2 + 2y = (y + 1)^2 – 1 \)
Substitute:
\( \left( x + \frac{p}{2} \right)^2 – \left( \frac{p}{2} \right)^2 + (y + 1)^2 – 1 + q = 0 \)
\( \left( x + \frac{p}{2} \right)^2 + (y + 1)^2 = \frac{p^2}{4} + 1 – q \)
Rewrite:
\( \left( x – \left( -\frac{p}{2} \right) \right)^2 + (y – (-1))^2 = \frac{p^2}{4} + 1 – q \)
Answer: \( \left( x – \left( -\frac{p}{2} \right) \right)^2 + (y – (-1))^2 = \frac{p^2}{4} + 1 – q \), where \( a = -\frac{p}{2} \), \( r^2 = \frac{p^2}{4} + 1 – q \)
Part (b)(i): Equation of the normal at \( A(4,3) \)
Tangent: \( x + 2y = 10 \), or \( y = -\frac{1}{2}x + 5 \)
Gradient of tangent: \( -\frac{1}{2} \)
Gradient of normal (perpendicular): \( 2 \)
At \( A(4,3) \):
\( y – 3 = 2(x – 4) \)
\( y = 2x – 8 + 3 \)
Answer: \( y = 2x – 5 \)
Part (b)(ii): Find \( p \) and \( q \)
Normal \( y = 2x – 5 \) passes through center \( \left( -\frac{p}{2}, -1 \right) \):
\( -1 = 2 \left( -\frac{p}{2} \right) – 5 \)
\( -1 = -p – 5 \)
\( p = 4 \)
Distance from center to \( A(4,3) \) equals radius:
Center: \( \left( -\frac{4}{2}, -1 \right) = (2, -1) \)
\( r^2 = (4 – 2)^2 + (3 – (-1))^2 = 4 + 16 = 20 \)
\( r^2 = \frac{p^2}{4} + 1 – q \)
\( 20 = \frac{4^2}{4} + 1 – q \)
\( 20 = 4 + 1 – q \)
\( q = 5 – 20 = -15 \)
Answer: \( p = -4 \), \( q = -15 \)
▶️ Answer/Explanation
Solution
Part (a): Distance between centres
For \( C_1 \): \( x^2 + y^2 + 6x – 10y + 18 = 0 \)
Complete the square:
\( x^2 + 6x = (x + 3)^2 – 9 \)
\( y^2 – 10y = (y – 5)^2 – 25 \)
\( (x + 3)^2 – 9 + (y – 5)^2 – 25 + 18 = 0 \)
\( (x + 3)^2 + (y – 5)^2 = 16 \)
Center: \( (-3, 5) \), Radius: \( r_1 = 4 \)
For \( C_2 \): \( (x – 9)^2 + (y + 4)^2 – 64 = 0 \)
\( (x – 9)^2 + (y + 4)^2 = 64 \)
Center: \( (9, -4) \), Radius: \( r_2 = 8 \)
Distance:
\( d = \sqrt{(9 – (-3))^2 + (-4 – 5)^2} = \sqrt{12^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \)
Answer: \( 15 \)
Part (b): Greatest and least possible values of d
Distance between centers: \( d = 15 \)
Radii: \( r_1 = 4 \), \( r_2 = 8 \)
Greatest distance (P and Q on opposite sides):
\( d_{\max} = 15 + 4 + 8 = 27 \)
Least distance (P and Q on same side):
\( d_{\min} = 15 – (4 + 8) = 15 – 12 = 3 \)
Answer: Greatest \( d = 27 \), Least \( d = 3 \)
▶️ Answer/Explanation
Solution
Line: \( y = -3x – k \)
Substitute into curve \( x^2 – 3xy – 40 = 0 \):
\( x^2 – 3x(-3x – k) – 40 = 0 \)
\( x^2 + 9x^2 + 3kx – 40 = 10x^2 + 3kx – 40 = 0 \)
Discriminant: \( \Delta = (3k)^2 – 4 \cdot 10 \cdot (-40) = 9k^2 + 1600 \)
Since \( 9k^2 + 1600 > 0 \) for all \( k \), the quadratic has real roots.
Answer: The curve and line intersect for all \( k \).