Past Exam Questions

(a) The equation of a circle with center (h, k) and radius r is \((x-h)^2 + (y-k)^2 = r^2\). For circle P, the center is (0, 2) and the radius is 10, so the equation is \(x^2 + (y-2)^2 = 100\).

(b) The gradient of the radius at point A is \(\frac{10-2}{6-0} = \frac{4}{3}\). The gradient of the tangent is the negative reciprocal, \(-\frac{3}{4}\). Using point-slope form, the equation of the tangent at (6, 10) is \(y - 10 = -\frac{3}{4}(x - 6)\), which simplifies to \(y = \frac{3}{4}x + \frac{29}{2}\).

(c) The center of circle Q is where the tangent meets the y-axis, which is \(\left(0, \frac{29}{2}\right)\). The radius is \(\frac{5}{2} \sqrt{5}\), so the equation is \(x^2 + \left(y - \frac{29}{2}\right)^2 = \left(\frac{5}{2} \sqrt{5}\right)^2 = 31.25\). To verify the y-coordinates of intersection, solve \(x^2 + (11-2)^2 = 100\) and \(x^2 + (11-11)^2 = 31.25\), confirming y-coordinates are 11.

(d) Substitute the tangent equation into circle Q's equation: \(x^2 + \left(\frac{3}{4}x + \frac{29}{2} - \frac{29}{2}\right)^2 = 31.25\). Solving gives \(x = \pm \sqrt{20}\). For each x, find y: \(y = \frac{3}{4}x + \frac{29}{2}\), resulting in \(y = \frac{58 \pm 3\sqrt{20}}{4}\).

First, calculate the radius of the circle using point A and the center (-2, 5):

\(r^2 = (7 + 2)^2 + (12 - 5)^2 = 9^2 + 7^2 = 81 + 49 = 130\)

The equation of the circle is:

\((x + 2)^2 + (y - 5)^2 = 130\)

Substitute the line equation \(y = -2x + 26\) into the circle equation:

\((x + 2)^2 + (-2x + 26 - 5)^2 = 130\)

\((x + 2)^2 + (-2x + 21)^2 = 130\)

Expand and simplify:

\((x^2 + 4x + 4) + (4x^2 - 84x + 441) = 130\)

\(5x^2 - 80x + 445 = 130\)

\(5x^2 - 80x + 315 = 0\)

Factorize:

\(5(x - 9)(x - 7) = 0\)

Thus, \(x = 9\) or \(x = 7\).

For \(x = 9\), substitute back into the line equation:

\(y = -2(9) + 26 = -18 + 26 = 8\)

Therefore, the coordinates of B are \((9, 8)\).

(a) To find the value of p such that AC and BC are equal in length, we use the distance formula:

\((5 - 2p)^2 + (p + 2)^2 = (10 - 2p)^2 + (3 - p)^2\)

Expanding both sides:

\(25 - 20p + 4p^2 + p^2 + 4p + 4 = 100 - 40p + 4p^2 + 9 - 6p + p^2\)

Simplifying gives:

\(30p = 80\)

Thus, \(p = \frac{8}{3}\).

(b)(i) Given that AC is perpendicular to BC, the slopes \(m_{AC}\) and \(m_{BC}\) must satisfy \(m_{AC} \times m_{BC} = -1\).

\(m_{AC} = \frac{p + 2}{2p - 5}\) and \(m_{BC} = \frac{p - 3}{2p - 10}\)

Setting \(\frac{p + 2}{2p - 5} \times \frac{p - 3}{2p - 10} = -1\) and solving gives:

\(p^2 - p - 6 = 0\)

Factoring gives \((p - 4)(p - 11) = 0\), so \(p = 4\) (ignoring \(p = \frac{11}{5}\) as p is an integer).

(b)(ii) The circle passes through A, B, and C. The midpoint of AB is \((7.5, 0.5)\).

The radius squared \(r^2 = \frac{50}{4}\) or \(r = \frac{5\sqrt{2}}{2}\).

The equation of the circle is \((x - 7.5)^2 + (y - 0.5)^2 = \frac{50}{4}\).

Expanding and simplifying gives:

\(x^2 + y^2 - 15x - y + 44 = 0\).

(a) To find the perpendicular bisector of AB, first find the midpoint of AB:

Midpoint = \(\left( \frac{10+5}{2}, \frac{2+(-1)}{2} \right) = \left( \frac{15}{2}, \frac{1}{2} \right)\).

Next, find the gradient of AB:

Gradient of AB = \(\frac{-1 - 2}{10 - 5} = -\frac{3}{5}\).

The gradient of the perpendicular bisector is the negative reciprocal:

Gradient = \(\frac{5}{3}\).

Using the point-slope form, the equation of the perpendicular bisector is:

\(y - \frac{1}{2} = \frac{5}{3} \left( x - \frac{15}{2} \right)\).

Simplifying gives:

\(y = \frac{5}{3}x - 12\).

(b) The equation of a circle with center (h, k) and radius r is:

\((x-h)^2 + (y-k)^2 = r^2\).

Here, the center is A (5, 2) and it passes through B (10, -1), so:

Radius = \(\sqrt{(10-5)^2 + (-1-2)^2} = \sqrt{25 + 9} = \sqrt{34}\).

The equation of the circle is:

\((x-5)^2 + (y-2)^2 = 34\).

1. The gradient of line AB is \(\frac{q - 3}{p - 8}\).

2. The perpendicular bisector has a gradient of -2, so the gradient of AB is \(\frac{1}{2}\).

3. Set \(\frac{q - 3}{p - 8} = \frac{1}{2}\) and solve for q:

\(q - 3 = \frac{1}{2}(p - 8)\)

\(2q - p = -2\)

4. The midpoint of AB is \(\left( \frac{8 + p}{2}, \frac{3 + q}{2} \right)\).

5. Substitute the midpoint into the equation of the perpendicular bisector:

\(\frac{3 + q}{2} = -2 \left( \frac{8 + p}{2} \right) + 4\)

\(3 + q = -2(8 + p) + 8\)

\(3 + q = -16 - 2p + 8\)

\(q = -11 - 2p\)

6. Solve the system of equations:

\(2q - p = -2\)

\(q = -11 - 2p\)

Substitute \(q = -11 - 2p\) into \(2q - p = -2\):

\(2(-11 - 2p) - p = -2\)

\(-22 - 4p - p = -2\)

\(-5p = 20\)

\(p = -4\)

Substitute \(p = -4\) into \(q = -11 - 2p\):

\(q = -11 - 2(-4)\)

\(q = -11 + 8\)

\(q = -3\)

(i) To find the gradient of line \(AB\), use the formula:

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Substitute the coordinates of \(A(-3k - 1, k + 3)\) and \(B(k + 3, 3k + 5)\):

\(m = \frac{(3k + 5) - (k + 3)}{(k + 3) - (-3k - 1)} = \frac{2k + 2}{4k + 4}\)

Simplify:

\(m = \frac{2(k + 1)}{4(k + 1)} = \frac{1}{2}\)

The gradient is \(\frac{1}{2}\), independent of \(k\).

(ii) To find the equation of the perpendicular bisector, first find the midpoint of \(AB\):

\(\text{Midpoint} = \left( \frac{-3k - 1 + k + 3}{2}, \frac{k + 3 + 3k + 5}{2} \right) = \left( \frac{-2k + 2}{2}, \frac{4k + 8}{2} \right)\)

Simplify:

\(\left( -k + 1, 2k + 4 \right)\)

The gradient of the perpendicular bisector is the negative reciprocal of \(\frac{1}{2}\):

\(-2\)

Using the point-slope form of a line equation:

\(y - (2k + 4) = -2(x - (-k + 1))\)

Simplify:

\(y - 2k - 4 = -2(x + k - 1)\) \(y - 2k - 4 = -2x - 2k + 2\) \(y + 2x = 6\)

1. Find the gradient of line \(AB\):

\(\text{Gradient of } AB = \frac{5h - h}{4h + 6 - h} = \frac{4h}{3h + 6}\)

2. The gradient of the perpendicular bisector is \(-\frac{3}{2}\), so:

\(\frac{4h}{3h + 6} \times \left(-\frac{3}{2}\right) = -1\)

3. Solve for \(h\):

\(\frac{4h}{3h + 6} = \frac{2}{3}\)

\(12h = 2(3h + 6)\)

\(12h = 6h + 12\)

\(6h = 12\)

\(h = 2\)

4. Find the midpoint of \(AB\):

\(\left(\frac{5h + 6}{2}, \frac{3h}{2}\right) = (8, 6)\)

5. Substitute the midpoint into the equation of the bisector:

\(3(8) + 2(6) = k\)

\(24 + 12 = k\)

\(k = 36\)

(i) To find the intersection points, set the equations equal:

\(\frac{1}{x} + c = cx - 3\)

Multiply through by \(x\) to clear the fraction:

\(1 + cx = cx^2 - 3x\)

Rearrange to form a quadratic equation:

\(cx^2 - 3x - cx + 1 = 0\)

\(cx^2 - cx - 3x + 1 = 0\)

Use the quadratic formula \(b^2 - 4ac\) to find the discriminant:

\((c + 3)^2 - 4c = c^2 + 6c + 9 - 4c = c^2 + 2c + 9\)

For real solutions, the discriminant must be non-negative:

\(c^2 + 2c + 9 \geq 0\)

Critical values are \(c = -1\) and \(c = -9\). Thus, \(c \leq -9\) or \(c \geq -1\).

(ii) For tangency, the discriminant must be zero. Substitute \(c = -1\) and \(c = -9\) into the quadratic equation:

For \(c = -1\):

\(-x^2 - 2x - 1 = 0\)

Solving gives \(x = -1\).

For \(c = -9\):

\(-9x^2 + 6x - 1 = 0\)

Solving gives \(x = \frac{1}{3}\).

Part (i):

1. Find the midpoint of AB:

\(\left( \frac{1+5}{2}, \frac{1+9}{2} \right) = (3, 5)\)

2. Calculate the gradient of AB:

\(\frac{9-1}{5-1} = 2\)

3. The gradient of the perpendicular bisector is the negative reciprocal:

\(-\frac{1}{2}\)

4. Use the point-slope form to find the equation:

\(y - 5 = -\frac{1}{2}(x - 3)\)

5. Simplify to get:

\(2y = 13 - x\)

Part (ii):

1. Substitute the equation of the perpendicular bisector into the curve equation:

\(6\left( \frac{13-x}{2} \right) = 5x^2 - 18x + 19\)

2. Simplify and solve the quadratic equation:

\(5x^2 - 3x - 4 = 0\)

3. Factorize to find x:

\((x-4)(x+1) = 0\)

4. Solutions are x = 4 and x = -1.

5. Find corresponding y values:

For x = 4, \(y = 4.5\); for x = -1, \(y = 7\).

6. Calculate the distance CD:

\(CD^2 = (4 - (-1))^2 + (4.5 - 7)^2 = 5^2 + (-4.5)^2 = 25 + 6.25 = 31.25\)

7. Therefore, \(CD = \sqrt{\frac{125}{4}}\).

(i) The gradient of line \(AP\) is given by:

\(\frac{b - 1}{a + 1} = 2\)

Solving for \(b\):

\(b - 1 = 2(a + 1)\) \(b - 1 = 2a + 2\) \(b = 2a + 3\)

Thus, \(b = 2a + 3\).

(ii) The distance \(AB\) is:

\(AB = \sqrt{(10 - (-1))^2 + (-1 - 1)^2} = \sqrt{11^2 + 2^2} = \sqrt{125}\)

Since \(AP = AB\), we have:

\(\sqrt{(a + 1)^2 + (b - 1)^2} = \sqrt{125}\)

Substitute \(b = 2a + 3\):

\((a + 1)^2 + (2a + 3 - 1)^2 = 125\) \((a + 1)^2 + (2a + 2)^2 = 125\) \((a + 1)^2 + 4(a + 1)^2 = 125\) \(5(a + 1)^2 = 125\) \((a + 1)^2 = 25\) \(a + 1 = 5 \quad \text{or} \quad a + 1 = -5\) \(a = 4 \quad \text{or} \quad a = -6\)

For \(a = 4\):

\(b = 2(4) + 3 = 11\)

For \(a = -6\):

\(b = 2(-6) + 3 = -9\)

Thus, the possible coordinates of \(P\) are \((4, 11)\) and \((-6, -9)\).

Step-by-step Solution:

(i) Find the equation of line AB:

The gradient of the perpendicular bisector is \(\frac{3}{2}\). Therefore, the gradient of line AB is the negative reciprocal, \(-\frac{2}{3}\).

Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the gradient and \((x_1, y_1)\) is a point on the line:

\(y - 6 = -\frac{2}{3}(x + 2)\)

Simplifying:

\(y - 6 = -\frac{2}{3}x - \frac{4}{3}\)

\(y = -\frac{2}{3}x + \frac{14}{3}\)

Multiplying through by 3 to eliminate fractions:

\(3y = -2x + 14\)

Rearranging gives:

\(3y + 2x = 14\)

(ii) Find the coordinates of point B:

The midpoint of AB is the point where the perpendicular bisector intersects AB. Solving the equations \(2y = 3x + 5\) and \(3y + 2x = 14\) simultaneously:

From \(2y = 3x + 5\), express \(y\):

\(y = \frac{3}{2}x + \frac{5}{2}\)

Substitute into \(3y + 2x = 14\):

\(3\left(\frac{3}{2}x + \frac{5}{2}\right) + 2x = 14\)

\(\frac{9}{2}x + \frac{15}{2} + 2x = 14\)

\(\frac{13}{2}x + \frac{15}{2} = 14\)

\(\frac{13}{2}x = \frac{13}{2}\)

\(x = 1\)

Substitute \(x = 1\) back into \(y = \frac{3}{2}x + \frac{5}{2}\):

\(y = \frac{3}{2}(1) + \frac{5}{2} = 4\)

The midpoint is \((1, 4)\). Since A is \((-2, 6)\), use the midpoint formula to find B:

\(\left(\frac{-2 + x}{2}, \frac{6 + y}{2}\right) = (1, 4)\)

\(\frac{-2 + x}{2} = 1 \Rightarrow x = 4\)

\(\frac{6 + y}{2} = 4 \Rightarrow y = 2\)

Thus, the coordinates of B are \((4, 2)\).

(i) Since B is the midpoint of AC, we have:

\(\frac{2 + x}{2} = n \quad \Rightarrow \quad x = 2n - 2\)

\(\frac{m + y}{2} = -6 \quad \Rightarrow \quad y = -12 - m\)

Thus, the coordinates of C are \((2n - 2, -12 - m)\).

(ii) The line \(y = x + 1\) passes through C, so:

\(-12 - m = (2n - 2) + 1\)

\(-12 - m = 2n - 1\)

\(m + 2n = -11\)

The line is perpendicular to AB, so the slope of AB is the negative reciprocal of 1:

\(\frac{m + 6}{2 - n} = -1\)

\(m + 6 = -2 + n\)

\(m - n = -8\)

Solving the equations:

\(m + 2n = -11\)

\(m - n = -8\)

Subtract the second from the first:

\(3n = -3\)

\(n = -1\)

Substitute \(n = -1\) into \(m - n = -8\):

\(m + 1 = -8\)

\(m = -9\)

Thus, \(m = -9\) and \(n = -1\).

1. The line \(\frac{x}{a} + \frac{y}{b} = 1\) intersects the x-axis at \(A(a, 0)\) and the y-axis at \(B(0, b)\).

2. The distance \(AB\) is given by the Pythagorean theorem:

\(AB = \sqrt{a^2 + b^2} = 10\)

\(a^2 + b^2 = 100\)

3. The mid-point \(M\) of \(AB\) is \(\left( \frac{a}{2}, \frac{b}{2} \right)\).

4. \(M\) lies on the line \(2x + y = 10\):

\(2 \left( \frac{a}{2} \right) + \frac{b}{2} = 10\)

\(a + \frac{b}{2} = 10\)

5. Solve the system of equations:

From \(a + \frac{b}{2} = 10\):

\(b = 20 - 2a\)

Substitute into \(a^2 + b^2 = 100\):

\(a^2 + (20 - 2a)^2 = 100\)

\(a^2 + 400 - 80a + 4a^2 = 100\)

\(5a^2 - 80a + 300 = 0\)

\(a^2 - 16a + 60 = 0\)

6. Solve the quadratic equation:

\((a - 6)(a - 10) = 0\)

\(a = 6 \text{ or } a = 10\)

7. If \(a = 6\), then \(b = 20 - 2 \times 6 = 8\).

8. If \(a = 10\), then \(b = 0\), which is not possible as \(b\) must be positive.

9. Therefore, \(a = 6\) and \(b = 8\).

First, find the midpoint C of line AB:

\(C = \left( \frac{14 + (-6)}{2}, \frac{-7 + 3}{2} \right) = (4, -2)\)

Next, find the slope of AB:

\(m_{AB} = \frac{3 - (-7)}{-6 - 14} = \frac{10}{-20} = -\frac{1}{2}\)

The slope of line CD, being perpendicular to AB, is the negative reciprocal:

\(m_{CD} = 2\)

Using point C(4, -2) in the point-slope form:

\(y + 2 = 2(x - 4)\)

Simplifying to slope-intercept form:

\(y = 2x - 8 - 2\)

\(y = 2x - 10\)

Since the line crosses the y-axis at D, set x = 0:

\(y = 2(0) - 10 = -10\)

So, the equation of CD is:

\(y = 2x - 10\)

For the distance AD, use the distance formula:

\(AD = \sqrt{(14 - 4)^2 + (-7 - (-2))^2}\)

\(AD = \sqrt{10^2 + (-5)^2}\)

\(AD = \sqrt{100 + 25}\)

\(AD = \sqrt{125} = 5\sqrt{5}\)

Therefore, the distance AD is:

\(AD = 5\sqrt{5}\)

(i) To show that triangle ABC is isosceles, calculate the lengths of AB, BC, and AC:

\(AB^2 = (4 + 2)^2 + (6 + 1)^2 = 6^2 + 7^2 = 85\)

\(BC^2 = (6 - 4)^2 + (-3 - 6)^2 = 2^2 + 9^2 = 85\)

\(AC^2 = (6 + 2)^2 + (-3 + 1)^2 = 8^2 + 2^2 = 68\)

Since \(AB = BC\), triangle ABC is isosceles.

To find the area of triangle ABC, use the formula:

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

Let M be the midpoint of AC, \(M = (2, -2)\).

Calculate \(BM\):

\(BM = \sqrt{(4 - 2)^2 + (6 + 2)^2} = \sqrt{2^2 + 8^2} = \sqrt{68}\)

Area \(\triangle ABC = \frac{1}{2} \times \sqrt{68} \times \sqrt{68} = 34\).

(ii) Find the gradient of AB:

\(\text{Gradient of } AB = \frac{6 - (-1)}{4 - (-2)} = \frac{7}{6}\)

Equation of AB:

\(y + 1 = \frac{7}{6}(x + 2)\)

Find the gradient of CD (perpendicular to AB):

\(\text{Gradient of } CD = -\frac{6}{7}\)

Equation of CD:

\(y + 3 = -\frac{6}{7}(x - 6)\)

Solving the equations:

\(2 = -\frac{6}{7}x + \frac{36}{7}\)

\(\frac{7}{6}x - \frac{14}{6}\)

\(x = \frac{34}{85} = \frac{2}{5}\)

(a) To find \(k\), substitute \(x = 2\) into the curve equation:

\(y = (2 - 3)\sqrt{2 + 1} + 3\)

\(y = (-1)\sqrt{3} + 3\)

\(y = 3 - \sqrt{3}\)

\(k = 1.2679\)

(b) The gradient of \(AE\) is given by:

\(\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1}\)

\(\text{Gradient} = \frac{3 - 1.2679}{3 - 2}\)

\(\text{Gradient} = 1.7321\)

(c) The gradients of \(BE, CE, DE\) are approaching 2 as \(x\) approaches 3. This suggests that the gradient of the curve at \(E\) is the limit of these gradients, which is approximately 2.

Part (i):

The slope of line \(AB\) is \(m = \frac{3 - 7}{8 - 0} = -\frac{1}{2}\).

The equation of line \(AB\) is \(y = -\frac{1}{2}x + 7\).

Substitute \(C(3k, k)\) into the line equation:

\(k = -\frac{1}{2}(3k) + 7\)

\(k = -\frac{3}{2}k + 7\)

\(2k = -3k + 14\)

\(5k = 14\)

\(k = 2.8\)

Part (ii):

The midpoint of \(AB\) is \(M(4, 5)\).

The perpendicular slope is \(2\), so the equation of the perpendicular bisector is:

\(y - 5 = 2(x - 4)\)

\(y = 2x - 3\)

Substitute \(C(3k, k)\) into the bisector equation:

\(k = 2(3k) - 3\)

\(k = 6k - 3\)

\(5k = 3\)

\(k = 0.6\)

(i) To find the equation of the perpendicular bisector of \(AB\):

1. Calculate the midpoint of \(AB\):

\(\left( \frac{5+9}{2}, \frac{7+(-1)}{2} \right) = (7, 3)\)

2. Find the slope of \(AB\):

\(m = \frac{-1 - 7}{9 - 5} = -2\)

3. The slope of the perpendicular bisector is the negative reciprocal:

\(m_{\text{perp}} = \frac{1}{2}\)

4. Use the point-slope form to find the equation:

\(y - 3 = \frac{1}{2}(x - 7)\)

(ii) To find the distance \(BX\):

1. Find the equation of the line through \(C(1, 2)\) parallel to \(AB\):

\(y - 2 = -2(x - 1)\)

2. Solve the system of equations:

\(\begin{align*} y - 3 &= \frac{1}{2}(x - 7) \\ y - 2 &= -2(x - 1) \end{align*}\)

3. Substitute and solve:

\(\frac{1}{2}x - \frac{1}{2} = -2x + 4\)

\(x = \frac{9}{5}, \quad y = \frac{2}{5}\)

4. Calculate the distance \(BX\):

\(BX = \sqrt{(9 - \frac{9}{5})^2 + (-1 - \frac{2}{5})^2}\)

\(BX = \sqrt{7.2^2 + 1.4^2}\)

\(BX \approx 7.33\)

(i) Calculate the distance AB:

\(AB = \sqrt{(5 - (-3))^2 + (1 - 7)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = 10\)

Calculate the distance BC:

\(BC = \sqrt{(5 - (-1))^2 + (1 - k)^2} = \sqrt{6^2 + (k - 1)^2}\)

Given AB = BC, set the equations equal:

\(10 = \sqrt{36 + (k - 1)^2}\)

Square both sides:

\(100 = 36 + (k - 1)^2\)

Simplify:

\(64 = (k - 1)^2\)

Take the square root:

\(k - 1 = \pm 8\)

So, \(k = 9\) or \(k = -7\).

(ii) Find the midpoint M of AB:

\(M = \left( \frac{-3 + 5}{2}, \frac{7 + 1}{2} \right) = (1, 4)\)

Calculate the slope of AB:

\(m_{AB} = \frac{1 - 7}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4}\)

The slope of the perpendicular bisector is the negative reciprocal:

\(m_{\text{perp}} = \frac{4}{3}\)

Equation of the perpendicular bisector:

\(y - 4 = \frac{4}{3}(x - 1)\)

Set \(y = 0\) to find the x-intercept:

\(0 - 4 = \frac{4}{3}(x - 1)\) \(-4 = \frac{4}{3}(x - 1)\) \(-12 = 4(x - 1)\) \(-12 = 4x - 4\) \(-8 = 4x\) \(x = -2\)

Thus, the coordinates of D are \((-2, 0)\).