Past Exam Questions

(a) The equation of a circle with center (h, k) and radius r is given by \((x - h)^2 + (y - k)^2 = r^2\). Here, the center is (5, 2) and the circle passes through (7, 5).

Calculate the radius: \(r^2 = (7 - 5)^2 + (5 - 2)^2 = 2^2 + 3^2 = 4 + 9 = 13\).

Thus, the equation of the circle is \((x - 5)^2 + (y - 2)^2 = 13\).

(b) Substitute \(y = 5x - 10\) into the circle's equation: \((x - 5)^2 + ((5x - 10) - 2)^2 = 13\).

Simplify: \((x - 5)^2 + (5x - 12)^2 = 13\).

Expand and simplify: \(26x^2 - 130x + 156 = 0\).

Factorize: \(26(x - 2)(x - 3) = 0\).

Solutions are \(x = 2\) and \(x = 3\).

Find corresponding y-values: \(y = 5(2) - 10 = 0\) and \(y = 5(3) - 10 = 5\).

Points of intersection are (2, 0) and (3, 5).

Length of chord AB: \(AB = \sqrt{(3 - 2)^2 + (5 - 0)^2} = \sqrt{1 + 25} = \sqrt{26}\).

(a) Calculate the gradients of \(AB\) and \(BC\):

Gradient of \(AB = \frac{-3}{5}\), Gradient of \(BC = \frac{5}{3}\).

The product of the gradients \(m_{AB} \times m_{BC} = -1\), confirming \(\angle ABC = 90^\circ\).

(b) The center \(D\) is the midpoint of \(AC\):

\(D = \left( \frac{-2+6}{2}, \frac{3+5}{2} \right) = (2, 4)\).

(c) Use the circle equation \((x - x_c)^2 + (y - y_c)^2 = r^2\):

\((x-2)^2 + (y-4)^2 = 17\).

(d) Find \(E\) such that \(BE\) is a diameter:

\(E = (1, 8)\).

The gradient of the tangent is the negative reciprocal of \(BE\):

Gradient of \(BE = -4\), so gradient of tangent = \(\frac{1}{4}\).

Equation of the tangent: \(y - 8 = \frac{1}{4}(x - 1)\).

(a) To show that line l is tangent to the circle at A, we need to prove that the distance from the center of the circle to the line is equal to the radius of the circle.

The equation of the circle is \((x - 5)^2 + (y - 11)^2 = 52\).

The equation of the line is \(y - 5 = -\frac{2}{3}(x - 1)\), which simplifies to \(y = -\frac{2}{3}x + \frac{17}{3}\).

Substitute \((1, 5)\) into the circle equation: \((5 - 1)^2 + (11 - 5)^2 = 52\), which holds true, confirming A is on the circle.

The gradient of the radius at A is the negative reciprocal of the line's gradient, \(\frac{3}{2}\), confirming the line is tangent.

(b) For the other circle, the center must be such that the line is tangent at A. The center is \((-3, -1)\) because the line's perpendicular distance to this center is the radius \(\sqrt{52}\).

The equation of the circle is \((x + 3)^2 + (y + 1)^2 = 52\).

(a) To find p, use the condition that line AB is perpendicular to line BC. The gradient of AB is \(\frac{7 - 4}{p - 6}\) and the gradient of BC is \(\frac{18 - 7}{14 - p}\). Since they are perpendicular, their product is -1:

\(\frac{7 - 4}{p - 6} \times \frac{18 - 7}{14 - p} = -1\)

Solving this equation gives \(p^2 - 20p + 51 = 0\). Solving the quadratic equation, we find \(p = 3\) or \(p = 17\). Given \(p < 10\), the value of \(p\) is 3.

(b) The center of the circle is the midpoint of AC, which is \((10, 11)\). The radius is the distance from the center to any of the points, say A: \(\sqrt{(10 - 6)^2 + (11 - 4)^2} = \sqrt{65}\). Thus, the equation of the circle is \((x - 10)^2 + (y - 11)^2 = 65\).

(c) The gradient of the radius at C is \(\frac{11 - 18}{10 - 14} = \frac{7}{4}\). The tangent at C is perpendicular to this radius, so its gradient is \(-\frac{4}{7}\). Using point-slope form at C (14, 18), the equation is:

\(y - 18 = -\frac{4}{7}(x - 14)\)

Rearranging gives \(4x + 7y - 182 = 0\).

(a) To find the \(x\)-coordinates where the circle intersects the \(x\)-axis, set \(y = 0\) in the circle's equation:

\(x^2 - 4x - 77 = 0\).

Factor the quadratic equation:

\((x + 7)(x - 11) = 0\).

Thus, the \(x\)-coordinates are \(-7\) and \(11\).

(b) The center of the circle \(C\) is \((2, -3)\). The gradient of \(AC\) is \(\frac{1}{3}\) or the gradient of \(BC\) is \(\frac{1}{3}\).

The gradient of the tangent at \(A\) is \(3\) or the gradient of the tangent at \(B\) is \(-3\).

The equations of the tangents are \(y = 3x + 21\) and \(y = -3x + 33\).

Set the equations equal to find the intersection:

\(3x + 21 = -3x + 33\).

Solve for \(x\):

\(6x = 12\)

\(x = 2\).

Substitute \(x = 2\) into one of the tangent equations to find \(y\):

\(y = 3(2) + 21 = 27\).

Thus, the point of intersection is \((2, 27)\).

(a) To find the equation of the circle, we first determine the center. The perpendicular bisectors of the chords \(AB\) and \(BC\) intersect at the center of the circle. The midpoint of \(AB\) is \((7, 5)\) and the midpoint of \(BC\) is \((4, 9)\). The center of the circle is \((4, 5)\).

The radius \(r\) is the distance from the center \((4, 5)\) to any of the points on the circle, such as \(A(7, 1)\):

\(r^2 = (7-4)^2 + (1-5)^2 = 3^2 + (-4)^2 = 9 + 16 = 25\)

Thus, the radius \(r = 5\).

The equation of the circle is \((x-4)^2 + (y-5)^2 = 25\).

(b) The gradient of the radius at \(B\) is:

\(\text{Gradient of radius} = \frac{9-5}{7-4} = \frac{4}{3}\)

The tangent at \(B\) is perpendicular to the radius, so its gradient is the negative reciprocal:

\(\text{Gradient of tangent} = -\frac{3}{4}\)

The equation of the tangent line at \(B(7, 9)\) is:

\(y - 9 = -\frac{3}{4}(x - 7)\)

Simplifying gives:

\(y = -\frac{3}{4}x + \frac{57}{4}\)

(a) Calculate the distance from \(T(-6, 6)\) to the center \(C(8, 4)\):

\((x - 8)^2 + (y - 4)^2 = (-6 - 8)^2 + (6 - 4)^2 = 200\)

\(\sqrt{200} = 10\), hence \(\sqrt{200} > 10\), so \(T\) is outside the circle.

(b) Use the formula for the angle between a tangent and a radius:

\(\text{angle} = \sin^{-1}\left(\frac{10}{10\sqrt{2}}\right) = 45^\circ\)

(c) Find the gradient \(m\) of \(CT\):

\(m = \frac{1}{7}\)

Equation of \(AB\) is \(y - 5 = 7(x - 1)\)

\(y = 7x - 2\)

(d) Substitute \(y = 7x - 2\) into the circle's equation:

\((x - 8)^2 + (7x - 2 - 4)^2 = 100\)

\(50x^2 - 100x = 0\)

\(x = 0\) and \(x = 2\)

(a) The radius \(r\) of the circle can be found using the distance formula between \(B(5, 1)\) and \(A(-1, -2)\):

\(r = \sqrt{(5 - (-1))^2 + (1 - (-2))^2} = \sqrt{6^2 + 3^2} = \sqrt{45}\)

The equation of the circle is \((x - 5)^2 + (y - 1)^2 = 45\).

(b) Point \(C\) is such that \(AC\) is a diameter, so \(C\) is at \((11, 4)\). The gradient of \(CD\) is \(-2\), which is perpendicular to the radius \(BC\) with gradient \(0.5\). Therefore, \(DC\) is a tangent.

(c) The other tangent from \(D\) touches the circle at \(E\). Using the equation of the tangent \(y = 2x + 6\), we find \(E(-1, 4)\).

(a) To find the equation of the tangent, first calculate the gradient of line AB:

\(m_{AB} = \frac{4 - 2}{-1 - 3} = -\frac{1}{2}\).

The gradient of the tangent is the negative reciprocal: \(2\).

Using point B (3, 2), the equation of the tangent is:

\(y - 2 = 2(x - 3)\).

(b) The radius of the circle with centre B is the distance AB:

\(AB^2 = (4 - 2)^2 + (-1 - 3)^2 = 20\).

The equation of the circle is:

\((x - 3)^2 + (y - 2)^2 = 20\).

(c) Substitute the equation of the tangent into the circle's equation:

\((x - 3)^2 + (2x - 6)^2 = 20\).

Expanding and simplifying:

\(5x^2 - 30x + 25 = 20\).

\(5(x^2 - 6x + 1) = 0\).

\((x - 5)(x - 1) = 0\).

Thus, \(x = 5\) or \(x = 1\).

(a) To find the perpendicular bisector, first find the midpoint of \(AB\). The midpoint is \((-1, 7)\).

Next, calculate the gradient of \(AB\):

\(m = \frac{11 - 3}{5 + 7} = \frac{8}{12} = \frac{2}{3}\).

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

Using the point-slope form, the equation is:

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

Simplifying gives:

\(2(y - 7) = -3(x + 1)\)

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

\(3x + 2y = 11\).

(b) The center of the circle lies on the line \(12x - 5y = 70\) and the perpendicular bisector \(3x + 2y = 11\). Solving these simultaneously:

\(12x - 5y = 70\)

\(3x + 2y = 11\)

Solving gives \(x = 5\) and \(y = -2\).

The radius is the distance from the center \((5, -2)\) to either \((-7, 3)\) or \((5, 11)\):

\(r = \sqrt{(5 - (-7))^2 + (-2 - 3)^2} = \sqrt{12^2 + 5^2} = 13\).

The equation of the circle is:

\((x - 5)^2 + (y + 2)^2 = 169\).

(a) Rewrite the circle equation in standard form: \((x-4)^2 + (y+2)^2 = 16 + 4 + 5\). The centre is \((4, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) The gradient from P (1, 2) to C (4, -2) is \(-\frac{4}{3}\). The tangent at P has a gradient of \(\frac{3}{4}\). The equation is \(y - 2 = \frac{3}{4}(x - 1)\) or \(4y = 3x + 5\).

(c) Q has the same y-coordinate as P, so \(y = 2\). Q is as far to the right of C as P is to the left, so \(x = 7\). Thus, Q is (7, 2).

(d) The gradient of the tangent at Q is \(-\frac{3}{4}\). The equation of the tangent at Q is \(y - 2 = -\frac{3}{4}(x - 7)\) or \(4y + 3x = 29\). Solving \(4y = 3x + 5\) and \(4y + 3x = 29\) gives \(T\) as \(\left(4, \frac{17}{4}\right)\).

(a) The midpoint of AB is the center of the circle C. Calculate the midpoint: \(\left(\frac{-1+7}{2}, \frac{-2+4}{2}\right) = (3, 1)\).

The radius is half the distance of AB. Calculate the distance: \(\sqrt{(7 - (-1))^2 + (4 - (-2))^2} = \sqrt{64 + 36} = 10\). Thus, the radius is 5.

The equation of the circle is \((x-3)^2 + (y-1)^2 = 5^2 = 25\).

(b) The gradient of AB is \(\frac{4 - (-2)}{7 - (-1)} = \frac{6}{8} = \frac{3}{4}\). The gradient of the tangent T is the negative reciprocal: \(-\frac{4}{3}\).

The equation of the tangent at B is \(y - 4 = -\frac{4}{3}(x - 7)\). Simplifying gives \(3y + 4x = 40\).

(c) The reflection of the circle C in the line T has its center at \((11, 7)\) (since B is the midpoint of the line joining the centers).

The radius remains the same, so the new equation is \((x-11)^2 + (y-7)^2 = 25\).

(a) The center of circle \(C_1\) is the midpoint of the diameter with end-points \((-3, -5)\) and \((7, 3)\). The midpoint is \(\left( \frac{-3+7}{2}, \frac{-5+3}{2} \right) = (2, -1)\).

The radius \(r\) is half the distance between the end-points: \(r = \frac{1}{2} \sqrt{(-3-7)^2 + (-5-3)^2} = \frac{1}{2} \sqrt{100 + 64} = \frac{1}{2} \times 12 = 6\).

The equation of circle \(C_1\) is \((x-2)^2 + (y+1)^2 = 41\).

(b) Circle \(C_2\) is obtained by translating \(C_1\) by \(\begin{pmatrix} 8 \\ 4 \end{pmatrix}\). The center of \(C_2\) is \((2+8, -1+4) = (10, 3)\).

The equation of circle \(C_2\) is \((x-10)^2 + (y-3)^2 = 41\).

(c) The gradient \(m\) of the line joining the centers \((2, -1)\) and \((10, 3)\) is \(m = \frac{3 - (-1)}{10 - 2} = \frac{4}{8} = \frac{1}{2}\).

The line \(RS\) is perpendicular to this line, so its gradient is \(-2\). Using the midpoint \((6, 1)\), the equation of \(RS\) is \(y - 1 = -2(x - 6)\), which simplifies to \(y = -2x + 13\).

(d) Substitute \(y = -2x + 13\) into the equation of \(C_1\): \((x-2)^2 + (-2x+13+1)^2 = 41\).

Expanding and simplifying gives \(x^2 - 4x + 4 + 4x^2 - 40x + 100 = 41\).

This simplifies to \(5x^2 - 60x + 159 = 0\).

(a) The gradient of \(AB\) is \(-1\) because the tangents have gradient 1 and are perpendicular to the radius. The center of the circle is \((4, -1)\). Using the point-slope form, the equation of \(AB\) is \(y + 1 = -1(x - 4)\), which simplifies to \(y = -x + 3\).

(b) Substitute \(y = -x + 3\) into the circle equation \((x-4)^2 + (y+1)^2 = 40\):

\((x-4)^2 + (-x+3+1)^2 = 40\)

\(2(x-4)^2 = 40\) or \((x-4)^2 = 20\)

\(x = 4 \pm \sqrt{20}\)

Coordinates of \(A\) are \((4 - \sqrt{20}, -1 + \sqrt{20})\).

(c) The equation of the tangent at \(A\) is found using the point \((4 - \sqrt{20}, -1 + \sqrt{20})\) and gradient 1:

\(y - (-1 + \sqrt{20}) = 1(x - (4 - \sqrt{20}))\)

\(y = x - 5 + 4\sqrt{5}\)

(a) Substitute the line equation \(y = x - 9\) into the circle equation:

\((x - 1)^2 + (x - 9 + 4)^2 = 40\)

Simplify to:

\(x^2 - 6x - 7 = 0\)

Factorize to find \(x\):

\((x + 1)(x - 7) = 0\)

Thus, \(x = -1\) or \(x = 7\).

For \(x = -1\), \(y = -1 - 9 = -10\).

For \(x = 7\), \(y = 7 - 9 = -2\).

Coordinates of intersection are \((-1, -10)\) and \((7, -2)\).

(b) The midpoint \(C\) of \(AB\) is:

\(C = \left(\frac{-1 + 7}{2}, \frac{-10 + (-2)}{2}\right) = (3, -6)\)

The radius is:

\(\sqrt{(7 - (-1))^2 + (-2 - (-10))^2} / 2 = \sqrt{32}\)

The equation of the circle with center \(C(3, -6)\) and radius \(\sqrt{32}\) is:

\((x - 3)^2 + (y + 6)^2 = 32\)

(a) Substitute \(y = \frac{1}{2}x + 6\) into the circle's equation:

\((x-a)^2 + \left(\frac{1}{2}x + 6 - 3\right)^2 = 20\)

\((x-a)^2 + \left(\frac{1}{2}x + 3\right)^2 = 20\)

\(\frac{5}{4}x^2 + (3-2a)x + a^2 - 11 = 0\)

Using the quadratic formula, solve for \(a\):

\((3-2a)^2 - 4\cdot\frac{5}{4}(a^2-1) = 0\)

\(a = 4\) or \(a = -16\)

(b) For \(a = 4\), the center of the circle is \((4, 3)\) and \(P = (2, 7)\). The gradient of the normal is \(-2\).

The equation of the normal is:

\(y - 3 = -2(x - 4)\) or \(y - 7 = -2(x - 2)\)

(c) The tangents parallel to the normal have the same gradient \(-2\). Using the circle's equation:

\((x-4)^2 + \left(\frac{1}{2}x + 3\right)^2 = 20\)

\(\frac{5}{4}x^2 - 10x = 0\)

\(x = 0\) or \(x = 8\)

Coordinates are \((0, 1)\) and \((8, 5)\).

Equations of tangents:

\(x + y = 1\) and \(2x + y = 21\)

(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:

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